Quiz about Chemistry in Class 12

Question 1

Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential and biology. The values of standard redox potential (E^0) of two half cell reactions decide which way the raction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^0 (V with respect to normal hydrogen eletrode ) values. Using this dat obtain the correct explanations to Questions  I_2+2e^_ rarr2I^-E^0=0.54  Cl_2+2e^_ rarr2Cl^-E^0=1.36 Mn^(3+)+e^_ rarrMn^(2+)E^0=1.50Fe^(3+)+e^_ rarrFe^(2+)E^0=0.77O_2+4H^++4e^(-) rarr 2H_2OE^0=1.23 Among the following identifiy the correct statement.

Accepted Answer

NA

Question 2

Reducing agent is

Accepted Answer

NA

Question 3

Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential and biology. The values of standard redox potential (E^0) of two half cell reactions decide which way the raction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^0 (V with respect to normal hydrogen eletrode ) values. Using this dat obtain the correct explanations to Questions  I_2+2e^_ rarr2I^-E^0=0.54  Cl_2+2e^_ rarr2Cl^-E^0=1.36 Mn^(3+)+e^_ rarrMn^(2+)E^0=1.50Fe^(3+)+e^_ rarrFe^(2+)E^0=0.77O_2+4H^++4e^(-) rarr 2H_2OE^0=1.23While Fe^(3+) is stable Mn^(3+) is not stable in acid soltuion because

Accepted Answer

NA

Question 4

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values.  (I_(2)+2e^(-) rarr 2I^(-)E^(@)=0.54)(Cl_(2) +2e^(-) rarr 2Cl^(-)E^(@)=1.36)(Mn^(3+)+e^(-)rarr Mn^(2+)E^(@)=1.50)(Fe^(3+)+e^(-) rarr Fe^(2+)E^(@)=0.77)(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)OE^(@)=1.23)  Using these data obtain the correct explanation for the following questions.  While Fe^(3+) is stable Mn^(3+) is not stable in acid solution because

Accepted Answer

NA

Question 5

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values.  (I_(2)+2e^(-) rarr 2I^(-)E^(@)=0.54)(Cl_(2) +2e^(-) rarr 2Cl^(-)E^(@)=1.36)(Mn^(3+)+e^(-)rarr Mn^(2+)E^(@)=1.50)(Fe^(3+)+e^(-) rarr Fe^(2+)E^(@)=0.77)(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)OE^(@)=1.23)  Using these data obtain the correct explanation for the following questions.  Sodium fusion extract obtained from anline on treatment with iron (II) Sulphate and H_(2)SO_(4) in presence of air gives a prussian blue precipitate. Hence the blue colour is due to the formation of

Accepted Answer

NA

Question 6

Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned.  I_(2)+e^(-) to 2I^(-) E^(@)=0.54 V  Cl_(2)+2e^(-) to 2Cl^(-) E^(@)=1.36 V Mn^(2+)+2e^(-) to Mn E^(@)=1.50 V  Fe^(3+)+e^(-) to Fe^(2+) E^(@)=0.77 V  O_(2)+4H^(+)+4e^(-) to 2H_(2)O E^(@)=1.23 V.  Sodium fusion extract obtained from aniline on treatment with iron (II) sulphate and H_(2)SO_(4) in the presence of air gives a prussian blue precipitate. The blue colour is due to the formation of

Accepted Answer

NA

Question 7

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values.  (I_(2)+2e^(-) rarr 2I^(-)E^(@)=0.54)(Cl_(2) +2e^(-) rarr 2Cl^(-)E^(@)=1.36)(Mn^(3+)+e^(-)rarr Mn^(2+)E^(@)=1.50)(Fe^(3+)+e^(-) rarr Fe^(2+)E^(@)=0.77)(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)OE^(@)=1.23)  Using these data obtain the correct explanation for the following questions.  Among the following identify the correct statement

Accepted Answer

NA

Question 8

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  Amongst the following identify the species with an atom in +6 oxidation state

Accepted Answer

NA

Question 9

Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned.  I_(2)+e^(-) to 2I^(-) E^(@)=0.54 V  Cl_(2)+2e^(-) to 2Cl^(-) E^(@)=1.36 V Mn^(2+)+2e^(-) to Mn E^(@)=1.50 V  Fe^(3+)+e^(-) to Fe^(2+) E^(@)=0.77 V  O_(2)+4H^(+)+4e^(-) to 2H_(2)O E^(@)=1.23 V.  While Fe^(2+) ion is stable Mn^(2+) ion is not stable in acid solution because

Accepted Answer

NA

Question 10

Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned.  I_(2)+e^(-) to 2I^(-) E^(@)=0.54 V  Cl_(2)+2e^(-) to 2Cl^(-) E^(@)=1.36 V Mn^(2+)+2e^(-) to Mn E^(@)=1.50 V  Fe^(3+)+e^(-) to Fe^(2+) E^(@)=0.77 V  O_(2)+4H^(+)+4e^(-) to 2H_(2)O E^(@)=1.23 V.  Among the following identify the correct statement

Accepted Answer

NA

Question 11

Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respectto normal hydrogen electrode ) Using this data  I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 Cl_2 + 2e^(-) to 2Cl^(-) E=1.36V  Mn^(3+) + e^(-) to Mn^(2+) E^(@)= 1.50 Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V  O_2 +4H^(+) + 4e^(-) to 2H_2O E^(@) = 1.23  While Fe^(3+) is stable Mn^(3+) is not stable in acid solution because

Accepted Answer

NA

Question 12

Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respectto normal hydrogen electrode ) Using this data  I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 Cl_2 + 2e^(-) to 2Cl^(-) E=1.36V  Mn^(3+) + e^(-) to Mn^(2+) E^(@)= 1.50 Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V  O_2 +4H^(+) + 4e^(-) to 2H_2O E^(@) = 1.23  Among the following identify the correct statement

Accepted Answer

NA

Question 13

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  A compound contains atoms of three elements A B and C. If the oxidation number of A is +2 B is +5 and that of C is -2 the possible formula of the compound is

Accepted Answer

NA

Question 14

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  A mole of N_(2)H_(4) loses 10 moles of electrons to forn a new compound Y. Assuming that all the nitrogen appears in the new compound what is the oxidation rate of nitrogen in Y ?  (There is no change in the oxidation number of hydrogen)

Accepted Answer

NA

Question 15

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  When copper is treated with a certain concentration of nitric acid nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation xCu+uHNO_(3)rarrCu(NO_(3))_(2)+NO+NO_(2)+H_(2)O. The coefficients of x and y are

Accepted Answer

NA

Question 16

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  In the reaction 3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-).

Accepted Answer

NA

Question 17

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  For the redox reaction xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)rarrxy and z are

Accepted Answer

NA

Question 18

Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept reductants. In terms of electronic concept reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atomsions while reductants involve increase in the oxidation number of one of its atomsions.  Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atomsmolecules decomposition of substances displacement of metals of non metals and disproportionation of a particular species which may be metals non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method.  From the reaction M^(x+)+MnO_(4)^(-)rarrMO_(3)^(-)+Mn^(2+)+12O_(2) If one mole of MnO_(4)^(-) oxidizes 1.67 moles of M^(x+) to MO_(3)^(-)then the value of x in the reaction is

Accepted Answer

NA

Question 19

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions.  I_(2)+2e^(-)to2I^(-)E^(0)=0.54  Cl_(2)+2e^(-)to2Cl^(-)E^(0)=1.36  Mn^(3+)+e^(-)toMn^(2+)E^(0)=1.50  Fe^(3+)+e^(-)toFe^(2+)E^(0)=0.77  O_(2)+4H^(+)+4e^(-)to2H_(2)OE^(0)=1.23  Q. Among the following identify the correct statement

Accepted Answer

NA

Question 20

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions.  I_(2)+2e^(-)to2I^(-)E^(0)=0.54  Cl_(2)+2e^(-)to2Cl^(-)E^(0)=1.36  Mn^(3+)+e^(-)toMn^(2+)E^(0)=1.50  Fe^(3+)+e^(-)toFe^(2+)E^(0)=0.77  O_(2)+4H^(+)+4e^(-)to2H_(2)OE^(0)=1.23  Q. While Fe^(3+) is stable Mn^(3+) is not stable in acid solution because

Accepted Answer

NA

Question 21

Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions.  I_(2)+2e^(-)to2I^(-)E^(0)=0.54  Cl_(2)+2e^(-)to2Cl^(-)E^(0)=1.36  Mn^(3+)+e^(-)toMn^(2+)E^(0)=1.50  Fe^(3+)+e^(-)toFe^(2+)E^(0)=0.77  O_(2)+4H^(+)+4e^(-)to2H_(2)OE^(0)=1.23  Q. Sodium fusion extract obtained from aniline on treatment with iron (II) sulphate and H_(2)SO_(4) in presence of air gives a prussian blue precipitate. the blue colour is due to the formation of

Accepted Answer

NA

Question 22

Redox reactins play a pivoted role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below is a set of half-cell reactions (acidic medium) along with their E^(@) values with respect to normal hydrogen electrode. Using this data obtained the correct explanations to questions 15-16  I_(2)+2e^(-)to2I^(-)E^(@)=1.36V  Mn^(3+)+e^(-)toMn^(2+)E^(@)=1.50V  Fe^(3+)+e^(-)toFe^(2+)E^(@)=0.77V  O_(2)+4H^(+)+4e^(-)to2H_(2)OE^(@)=1.23V  Q. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is (1 faraday =96500C)

Accepted Answer

NA

Question 23

Redness of blood is because of the presence of

Accepted Answer

NA

Question 24

Redish brown (choclate) ppt are formed with

Accepted Answer

NA

Question 25

Red P is used in making

Accepted Answer

NA

Question 26

Reddish brown vapours formed when a chloride is heatedwith K_(2)Cr_(2)O_(7) and conc. H_(2)SO_(4) are due to the formation of …..........

Accepted Answer

NA

Question 27

Reddish brown gas is obtained when the following are treated with conc. H_(2)SO_(4)

Accepted Answer

NA

Question 28

Reddish brown colourationwhen neutral FeCl_(3) is added to the CH_(3)CO O^(Θ) aq solutionis due to the formation of (a) ____.

Accepted Answer

NA

Question 29

Reddis-brown gas is obtained when the following are treated with conc. H_(2)SO_(4) ?

Accepted Answer

NA

Question 30

Reddish- brown (chocolate) ppt. are formed with

Accepted Answer

NA

Question 31

Red precipitate underset(NH_(4)OH)overset(Cu_(2)Cl_(2))larrP(C_(5)H_(8))overset(Ozonolysis)rarr2-Methylpropanoic acid + compound (Q) structure of P can be-

Accepted Answer

NA

Question 32

Red phosphorus is..............reactive than white phosphorus as red phosphorous is..........and consists of ................of P_(4) units.

Accepted Answer

NA

Question 33

Red phosphorus is used in the manufacture of ............ and ............

Accepted Answer

NA

Question 34

Red phosphorus is chemically unreactive because

Accepted Answer

NA

Question 35

Red ink is prepared from

Accepted Answer

NA

Question 36

Red P is prepared from white P by heating in vacuum

Accepted Answer

NA

Question 37

Red Liquor is

Accepted Answer

NA

Question 38

Red lead is

Accepted Answer

NA

Question 39

Red how steel rod on suddenly immersing in water becomes

Accepted Answer

NA

Question 40

Red hot steel rod on suddenly immersing in water becomes.

Accepted Answer

NA

Question 41

Red hot iron absorbs SO_(2) giving the product

Accepted Answer

NA

Question 42

Red hot carbon will remove oxygen from the oxides XO and Yo but not from ZO.Ywillremove oxygen from XO. Use this evidence to deduce the order of activity of the three metals XY and Z putting the most reactive first.

Accepted Answer

NA

Question 43

Red hot carbon will remove oxygen from the oxide AO and BO but not from MO while B will remove oxygen from AO. The activity of metals A B and M in decreasing order is

Accepted Answer

NA

Question 44

Red hot carbon removes oxygen from the oxide XO and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to deduce the order of activity of the three metals X Y and Z putting the most active first.

Accepted Answer

NA

Question 45

Red hot carbon will remove oxygen from the oxides XO and YO but not form ZO . Y will remove oxygen from XO . Use this evidence to reduce the order of activity of the three metals XY and Z putting the most reactive first

Accepted Answer

NA

Question 46

Redcoloured nitroso- nitroalkaneis formedby the actionof followingon primarynitroalkane

Accepted Answer

NA

Question 47

Rectified spirit obtained by fermentation contains4.5% of water. So in order to remove it rectified spirit is mixed with suitable quantity of benzene and heated Benzene hepls because

Accepted Answer

NA

Question 48

Red coloured compound formed by ethyl alcohol with cerric ammonium nitrate is

Accepted Answer

NA

Question 49

Red colour complex ion formed on adding FeCl_(3) to sodium extract when N and S both are present in organic compound is

Accepted Answer

NA

Question 50

Red and yellow phosphorus are

Accepted Answer

NA

Explore topic-wise InterviewSolutions in Current Affairs.

Reducing agent is :

Redness of blood is because of the presence of :

Redish brown (choclate) ppt are formed with

Red P is used in making

Reddish brown vapours formed when a chloride is heatedwith K_(2)Cr_(2)O_(7) and conc. H_(2)SO_(4) are due to the formation of "….........".

Reddish brown gas is obtained when the following are treated with conc. H_(2)SO_(4)

Reddish brown colourationwhen neutral FeCl_(3) is added to the CH_(3)CO O^(Θ) aq solutionis due to the formation of (a) ____.

Reddis-brown gas is obtained when the following are treated with conc. H_(2)SO_(4) ?

Reddish- brown (chocolate) ppt. are formed with:

Red precipitate underset(NH_(4)OH)overset(Cu_(2)Cl_(2))larrP(C_(5)H_(8))overset("Ozonolysis")rarr2-Methylpropanoic acid + compound (Q) structure of P can be-

Red phosphorus is..............reactive than white phosphorus as red phosphorous is..........and consists of ................of P_(4) units.

Red phosphorus is used in the manufacture of ............ and ............

Red phosphorus is chemically unreactive because:

Red ink is prepared from

Red P is prepared from white P by heating in vacuum

Red Liquor is :

Red lead is :

Red how steel rod on suddenly immersing in water becomes

Red hot steel rod on suddenly immersing in water becomes.

Red hot iron absorbs SO_(2) giving the product

Red hot carbon will remove oxygen from the oxides XO and Yo but not from ZO.Ywillremove oxygen from XO. Use this evidence to deduce the order of activity of the three metals X,Y, and Z, putting the most reactive first.

Red hot carbon will remove oxygen from the oxide AO and BO but not from MO, while B will remove oxygen from AO. The activity of metals A, B and M in decreasing order is

Red hot carbon removes oxygen from the oxide XO and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to deduce the order of activity of the three metals X, Y and Z putting the most active first.

Red hot carbon will remove oxygen from the oxides XO and YO but not form ZO . Y will remove oxygen from XO . Use this evidence to reduce the order of activity of the three metals X,Y and Z putting the most reactive first :

Redcoloured nitroso- nitroalkaneis formedby the actionof followingon primarynitroalkane

Rectified spirit obtained by fermentation contains4.5% of water. So in order to remove it, rectified spirit is mixed with suitable quantity of benzene and heated, Benzene hepls because

Red coloured compound formed by ethyl alcohol with cerric ammonium nitrate is

Red colour complex ion formed on adding FeCl_(3) to sodium extract when N and S both are present in organic compound is

Red and yellow phosphorus are:

1.	Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential and biology. The values of standard redox potential (E^0) of two half cell reactions decide which way the raction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^0 (V with respect to normal hydrogen eletrode ) values. Using this dat obtain the correct explanations to Questions I_2+2e^_ rarr2I^-E^0=0.54 Cl_2+2e^_ rarr2Cl^-E^0=1.36 Mn^(3+)+e^_ rarrMn^(2+)E^0=1.50Fe^(3+)+e^_ rarrFe^(2+)E^0=0.77O_2+4H^++4e^(-) rarr 2H_2OE^0=1.23 Among the following, identifiy the correct statement.
Answer» Chloride ion is oxidised by `O_2` `FE^(2+)` is oxidised by IODINE Iodine ion oxidised by CHLORINE `MN^(2+)` is oxidised by chlorine Answer :C

2.	Reducing agent is :
Answer» `SNO` `SnO_2` `SnCl_2` `SnCl_4` ANSWER :C

3.	Redox reactions play a pivoted role in chemistry and biology. The values of standard redox potential and biology. The values of standard redox potential (E^0) of two half cell reactions decide which way the raction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E^0 (V with respect to normal hydrogen eletrode ) values. Using this dat obtain the correct explanations to Questions I_2+2e^_ rarr2I^-E^0=0.54 Cl_2+2e^_ rarr2Cl^-E^0=1.36 Mn^(3+)+e^_ rarrMn^(2+)E^0=1.50Fe^(3+)+e^_ rarrFe^(2+)E^0=0.77O_2+4H^++4e^(-) rarr 2H_2OE^0=1.23While Fe^(3+) is stable, Mn^(3+) is not stable in acid soltuion because
Answer» `O_2` OXIDISES `MN^(2+)` to `Mn^(3+)` `O_2` oxidises both `Mn^(2+)` to `Mn^(3+)` and `FE^(2+)` to `Fe^(3+)` `Fe^(3+)` oxidises `H_2O " to " O_2` `Mn^(3+)` oxidises `H_2O " to " O_2` Answer :D

4.	Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values. {:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):} Using these data, obtain the correct explanation for the following questions. While Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because
Answer» `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)` `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` and `FE^(2+)` to `Fe^(3+)` `Fe^(3+)` oxidises `H_(2)O` to `O_(2)` `Mn^(3+)` oxidises `H_(2)O` to `O_(2)` Solution :Calculate the EMF of all the cell. Only the EMF of the cell INVOLVING OXIDATION of `H_(2)O` to `O_(2)` by `Mn^(3+)` is +ve `{:(Mn^(3+)+e^(-) rarr Mn^(2+)"]"xx4, E^(@)=+1.50 V),(2H_(2)O rarr 4H^(+)+O_(2)+4e^(-), E^(@)=-1.23 V),(BAR(4 Mn^(3+)+2 H_(2)O rarr 4 Mn^(2+)+O_(2)+4 H^(+)", "E_("cell")^(@)=+0.27 V)):}` THUS, option (d) is corect.

5.	Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values. {:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):} Using these data, obtain the correct explanation for the following questions. Sodium fusion extract, obtained from anline, on treatment with iron (II) Sulphate and H_(2)SO_(4) in presence of air gives a prussian blue precipitate. Hence, the blue colour is due to the formation of
Answer» `Fe_(4)[FE(CN)_(6)]_(3)` `Fe_(3)[Fe(CN)_(6)]_(2)` `Fe_(4) [Fe(CN)_(6)]_(2)` `Fe_(3)[Fe(CN)_(6)]_(3)` Solution :`Na+C+N RARR NaCN` `Fe^(2+)+6 CN^(-) rarr [Fe(CN)_(6)]^(4-)` In presence of air, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions. `{:(Fe^(2+) rarr Fe^(3+) +e^(-) "]"xx4, E^(@)=-0.77 V),(O_(2)+4 H^(+)+ 4e^(-) + 4e^(-) rarr 2H_(2)O, E^(@)=+1.23 V),(bar(4 Fe^(2+)+4 H^(+)+O_(2) rarr 4 Fe^(3+) +2 H_(2)O", "E_("cell")^(@)= +0.46 V)):}` `Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ION to form ferric ferrocyanide which has PRUSSIAN blu colour. `4 Fe^(3+)+3[ Fe(CN)_(6)]^(4-) rarr underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))`

7.	Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is Daniel Cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) aong with their E^(@) (V with respect to normal hydrogen electrode) values. {:(I_(2)+2e^(-) rarr 2I^(-),E^(@)=0.54),(Cl_(2) +2e^(-) rarr 2Cl^(-),E^(@)=1.36),(Mn^(3+)+e^(-)rarr Mn^(2+),E^(@)=1.50),(Fe^(3+)+e^(-) rarr Fe^(2+),E^(@)=0.77),(O_(2)+4H^(+)+4e^(-) rarr 2H_(2)O,E^(@)=1.23):} Using these data, obtain the correct explanation for the following questions. Among the following, identify the correct statement
Answer» CHLORIDE ion is oxidised by `O_(2)` `Fe^(2+)` is oxidised by iodide Iodide ion is oxidised by chlorine `MN^(2+)` is oxidised by chlorine Solution :Calculate the EMF of all the cells. Only the EMF of the cell involving the oxidation of `I^(-)` ion by `Cl_(2)` is +ve. `{:(2I^(-) rarr I_(2)+2e^(-)", "E^(@)=-0.54 V),(Cl_(2)+2e^(-) rarr 2Cl^(-)", "E^(@)=+1.36 V),(bar(Cl_(2)+2I^(-) rarr 2Cl^(-)+I_(2), E_("cell")^(@)+0.82 V)):}` Thus, option (C) is correct.

8.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. Amongst the following, identify the species with an atom in +6 oxidation state
Answer» `MnO_(6)^(-)` `CR(CN)_(6)^(3-)` `MnF_(6)^(2-)` `CrO_(2)Cl_(2)` Solution :O.N. of Cr in `CrO_(2)Cl_(2)` (chromyl chloride) = X (say) `:. x +2(-2) +2(-1)=0` `rArr x = +6`.

10.	Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials (E^(@)) of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with E^(@) values. Using the data obtain the correct explanation to the questions that are mentioned. I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V " Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V" Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V" Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V" O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V". Among the following, identify the correct statement :
Answer» `Cl^(-)` ion is oxidised by `O_(2)` `Fe^(2+)`ion is oxidised by iodine. `I^(-)` ion is oxidised by chlorine. `Mn^(2+)` ion is oxidised by chlorine. Solution :(c ) `2I^(-)+Cl_(2) to 2Cl^(-)+l_(2)` `I^(-)`ions is oxidised by `Cl_(2)` because E.m.f. of the cell comes out to be POSITIVE. `E_(cell)^(@)=E_(Cl_(2)//2Cl^(-))^(@)-E_(2I^(-)//I_(2))^(@)` `=1.36-0.54=0.82" V"`

11.	Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respectto normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) ""E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@)= 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O"" E^(@) = 1.23 , While Fe^(3+) is stable , Mn^(3+) is not stable in acid solution because :
Answer» `O_2 ` OXIDISES `Mn^(2+)` to `Mn^(3+)` `O_2` oxidises both` Mn^(2+)` to `Mn^(3+)` and ` FE^(2+) " to " Fe^(3+)` `Fe^(3+) " oxidises " H_2O " to " O_2` `Mn^(3+)` oxidises ` H_2O " to " O_2 ` SOLUTION :`Mn^(+3)` oxidises `H_2O ` to `O_2`

13.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. A compound contains atoms of three elements A, B and C. If the oxidation number of A is +2, B is +5 and that of C is -2 the possible formula of the compound is
Answer» `A_(3)(BC_(4))_(2)` `A_(3)(B_(4)C)_(2)` `ABC_(2)` `A_(3)(BC_(3))_(2)` Solution :SUM of oxidation number of all the elements in the molecular formula of any compound is zero. Putting the O.N of all the elements in the molecular formula of the COMPOUNDS, we find that in `A_(3)(BC_(4))_(2)`, the sum of O.N. of all the elements is , `3 XX (+2) + 2 xx [5 + 4(-2)]=0`.

23.	Redness of blood is because of the presence of :
Answer» IRON in HAEME pigment Haemoglobin Copper in haeme pigment All Answer :A

25.	Red P is used in making
Answer» AIR freshners Red plastics Red DYES for plastics Safety match-striking surface Answer :D

34.	Red phosphorus is chemically unreactive because:
Answer» It does not CONTAIN P-P bonds It does not contain TETRAHEDRAL `P_4` molecules It does not catch fire in air even upto `4000^@` C It has a POLYMERIC structure Answer :D

35.	Red ink is prepared from
Answer» Aniline Congo RED Phenol EOSIN Solution :Eosin is Red DYES.

14.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. A mole of N_(2)H_(4) loses 10 moles of electrons to forn a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation rate of nitrogen in Y ? (There is no change in the oxidation number of hydrogen)
Answer» `-1` `-3` `+3` `+5` Solution :Total O.N of 2 nitrigen ATOMS in `N_(2)H_(4)` is -2 SINCE it loses 10 moles of ELECTRONS, thereforem the total O.N of two nitrogen atoms in Y increases by 10 i.e., the total O.N of the two introgen atoms in`Y = -4 + 10 = +6` `:.` O.N of each nitrogen in `Y = 6//2 = +3`.

15.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation xCu+uHNO_(3)rarrCu(NO_(3))_(2)+NO+NO_(2)+H_(2)O. The coefficients of x and y are
Answer» 2 and 3 2 and 6 1 and 3 3 and 8 Solution :Balanced equations for producing NO and `NO_(2)` respectively are `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`....(i) `Cu+4HNO_(3)rarrCu(NO_(3))_(2)+2NO_(2)+2H_(2)O` ....(ii) Adding EQN (i) and eqn (ii), `4Cu+12HNO_(3)rarr4Cu(NO_(3))_(2)+2NO_(2)+2NO+6H_(2)O` or `2Cu+6HNO_(3)rarr2Cu(NO_(3))_(2)+NO_(2)+NO+3H_(2)O` THUS, coefficients X and y of Cu and `HNO_(3)` are respectively 2 and 6.

16.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. In the reaction 3Br_(2)+6CO_(3)^(2-)+3H_(2)Orarr5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-).
Answer» Bromine is oxidized and carbonate is REDUCED Bromine is reduced and water is oxidized Bromine is neither reduced nor oxidized Bromine is both reduced and oxidized Solution :In this REACTION disproportionation of Br takes place from O.N = 0 (in `Br_(2)`) to -1 (in `Br^(-)`) and +5 (in `BrO_(3)^(-)`).

17.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. For the redox reaction xMnO_(4)^(-)+yC_(2)O_(4)^(2-)+zH^(+)rarrx,y and z are
Answer» `"2516"` `"1652"` `"5162"` `"2165"` Solution :The balanced redox REACTION is `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)rarr2Mn^(2+)+10CO_(2)+8H_(2)O` Thus the coefficients of `MnO_(4)^(-), C_(2)O_(4)^(2-) and H^(+)` respectively are 2,5 and 16.

18.	Redox reactions involve simultaneous reduction-oxidation reactions. The process of oxidation involves addition of oxygen or any other electronegative element or loss of hydrogen or any other electropositive element. The reverse of this process is called reduction. Reduction also involves addition of electrons or decrease in the oxidation number an atom or ion present in a substance.Substances which bring about oxidation of other substances are called oxidants while those which bring about the reduction of other substances are called reductants. In terms of electronic concept, reductants. In terms of electronic concept, reductants are electron donors while oxidants are electron acceptors. Oxidants also involve decrease in oxidation number of one of its atoms/ions while reductants involve increase in the oxidation number of one of its atoms/ions. Oxidation numbers are always while numbers and must be always calculated on the basis of their structures and never from their molecular formulae. Redox reactants may involve combination of atoms/molecules, decomposition of substances, displacement of metals of non metals and disproportionation of a particular species which may be metals, non-metals or ions. Redox reactions can be balanced both by oxidation number method as well as by ion electron method. From the reaction M^(x+)+MnO_(4)^(-)rarrMO_(3)^(-)+Mn^(2+)+1//2O_(2) If one mole of MnO_(4)^(-) oxidizes 1.67 moles of M^(x+) to MO_(3)^(-),then the value of x in the reaction is
Answer» 5 3 2 0 Solution :`overset(+7)(MnO_(4)^(-))+5e^(-)RARROVERSET(+2)(Mn)` SINCE 1 mole of `MnO_(4)^(-)` accepts 5 moles of electrons, therefore 5 moles of electron are lost by 1.67 moles of `M^(x+)` 1 mole of `M^(x+)` will LOSE electrons `= 5//1.67 =3` mol (approx) Since `M^(x+)` changes to `MO_(3)^(-)` (where O.N of M = 5) by accepting 3 electrons `:.` OXIDISING state of `M^(x+)` i.e., `x = +5 - 3 = +2` `overset(+2)(M^(2+))rarroverset(+5)(M)O_(3)^(-)+3e^(-)`.

19.	Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their E^(@) (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. I_(2)+2e^(-)to2I^(-)""E^(0)=0.54 Cl_(2)+2e^(-)to2Cl^(-)""E^(0)=1.36 Mn^(3+)+e^(-)toMn^(2+)""E^(0)=1.50 Fe^(3+)+e^(-)toFe^(2+)""E^(0)=0.77 O_(2)+4H^(+)+4e^(-)to2H_(2)O""E^(0)=1.23 Q. Among the following, identify the correct statement
Answer» Chloride ion oxidised `O_(2)` `FE^(2+)` is oxidised by iodine IODIDE ion is oxidised by chlorine `Mn^(2+)` is oxidised by chlorine Solution :`2I^(-)+Cl_(2)toI_(2)+2Cl^(-)` `E^(o)=E_(I^(-)//I_(2))^(o)+E_(Cl_(2)//CL^(-))^(o)` `=-0.54+1.36` `E^(o)=0.82V` `E^o` is positive HENCE iodide ion is oxidized by chlorine.

22.	Redox reactins play a pivoted role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below is a set of half-cell reactions (acidic medium) along with their E^(@) values with respect to normal hydrogen electrode. Using this data, obtained the correct explanations to questions 15-16 I_(2)+2e^(-)to2I^(-),E^(@)=1.36V Mn^(3+)+e^(-)toMn^(2+),E^(@)=1.50V Fe^(3+)+e^(-)toFe^(2+),E^(@)=0.77V O_(2)+4H^(+)+4e^(-)to2H_(2)O,E^(@)=1.23V Q. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of H_(2) gas at the cathode is: (1 faraday =96500C)
Answer» `9.65xx10^(4)sec` `19.3xx10^(4)sec` `28.95xx10^(4)sec` `38.6xx10^(4)sec` Solution :FARADAY LAW equivalents of `H_(2)` produced `=(Ixxt(sec))/(96500)` `0.01xx2=(10xx10^(3)XXT)/(96500)=96500xx2=t` `19.3xx10^(3)sec=t`

24.	Redish brown (choclate) ppt are formed with
Answer» `CU^(+2)` and `[Fe(CN)_(6)]^(-2)` `Ba^(+2)` and `SO_(4)^(-2)` `Pb^(+2)` and `I^(-)` `Ag^(+)` and `I^(-)` ANSWER :A

26.	Reddish brown vapours formed when a chloride is heatedwith K_(2)Cr_(2)O_(7) and conc. H_(2)SO_(4) are due to the formation of "….........".
Answer» ANSWER :chromyl CHLORIDE `(CrO_(2)Cl_(2))`

27.	Reddish brown gas is obtained when the following are treated with conc. H_(2)SO_(4)
Answer» Be `NO_(2)^(-)` `NO_(3)` `l^(-)` Solution :`Br^(-)` gives `Br_(2).NO_(2)^(-)` and `NO_(3)^(-)` gives `NO_(2)` gas which are reddish BROWN in colour. Hence, (A), (B) and (C) are the CORRECT answers.

28.	Reddish brown colourationwhen neutral FeCl_(3) is added to the CH_(3)CO O^(Θ) aq solutionis due to the formation of (a) ____.
Answer» Answer :a.`[Fe_(3)(OH)_(2)(CH_(2)COO)_(4)]^(THETA)`

29.	Reddis-brown gas is obtained when the following are treated with conc. H_(2)SO_(4) ?
Answer» `BR^(-)` `NO_(2)^(-)` `NO_(3)^(-)` `SO_(3)^(2-)` ANSWER :A::B::C

30.	Reddish- brown (chocolate) ppt. are formed with:
Answer» `PB^(2+) and I^(-)` `Ba^(2+) and SO_(4)^(2-)` `Cu^(2+)and FE (CN)_(6)^(4-)` None of these Answer :C

31.	Red precipitate underset(NH_(4)OH)overset(Cu_(2)Cl_(2))larrP(C_(5)H_(8))overset("Ozonolysis")rarr2-Methylpropanoic acid + compound (Q) structure of P can be-
Answer» `CH_(3)-CH_(2)-CH_(2)-C-=CH` `CH_(3)-UNDERSET(CH_(3))underset(\|)(CH)-C-=CH` `CH_(3)C=C-CH_(2)-CH_(3)` `CH_(3)-underset(CH_(3))underset(\|)(CH)-CH=CH_(2)` SOLUTION :N//A

32.	Red phosphorus is..............reactive than white phosphorus as red phosphorous is..........and consists of ................of P_(4) units.
Answer» SOLUTION :LESS, POLYMERIC, CHAINS

33.	Red phosphorus is used in the manufacture of ............ and ............
Answer» SOLUTION :MATCHES, HBR and HI

36.	Red P is prepared from white P by heating in vacuum
Answer» `256^@` C `500^@` C `50^@` C `100^@` C Answer :A

37.	Red Liquor is :
Answer» `AL(OH)_3` `(CH_3COO)_3 Al` `Al_2(CO_3)_3` `Al_2(SO_4)_3` ANSWER :B

38.	Red lead is :
Answer» PBO `Pb_3O_4` `PbO_2` `HGS` ANSWER :B

39.	Red how steel rod on suddenly immersing in water becomes
Answer» SOFT and malleable Hard and brittle Tough and ductile Fibrous Answer :B