3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 27 InterviewSolution

1301.	The molar conductance of ammonium hydroxide solution of concentration 0.1 M, 0.01M and 0.001 M are 3.6, 11.3 and `34.0 ohm^(-1)cm^(-2)mol^(-1)` respectively. Calculate the degree of dissociation of `NH_(4)OH` at these concentrations. Molar conductance at infinite dilution for `Nh_(4)OH` is 271.1 ohm^(-1)cm^(-1)mol^(-1)`.
Answer» Correct Answer - 0.0133, 0.0417, 0.1254 At 0.1 M concentraton , `" "(alpha)=(Lambda_(m)^(C ))/(Lambda_(m)^(oo))=((3.6" ohm"^(-1)cm^(2)mol^(-1)))/((271.1" ohm"^(-1)cm^(2)mol^(-1)))=0.0133` At 0.01 M concentration , `" "(alpha)=(Lambda_(m)^(C ))/(Lambda_(m)^(oo))=((11.3" ohm"^(-1)cm^(2)mol^(-1)))/((271.1" ohm"^(-1)cm^(2)mol^(-1)))=0.0417` At 0.001 M concentration , `" "(alpha)=(Lambda_(m)^(C ))/(Lambda_(m)^(oo))=((34.0" ohm"^(-1)cm^(2)mol^(-1)))/((271.1" ohm"^(-1)cm^(2)mol^(-1)))=0.1254`.

Discussion

1302.	Statement-I: `H_(2)+O_(2)` fuel cell gives a constant voltage throughout its life. Because Statement-II: In this fuel cell, `H_(2)` reacts with `OH^(-)` ions yet the overall concentration of `OH^(-)` ions does not change.
Answer» Correct Answer - A

Discussion

1303.	Statement-I: `H_(2)+O_(2)` fuel cell gives a constant voltage throughout its life. Because Statement-II: In this fuel cell, `H_(2)` reacts with `OH^(-)` ions yet the overall concentration of `OH^(-)` ions does not change.A. If both (A) and (R) are correct, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - A

Discussion

1304.	Assertion `(A) : ` The presence of `CO_(2)` in the air accelerates corrosion. Reason `(R): CO_(2)` is a poisonous gas.A. If both (A) and (R) are correct, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct.
Answer» Correct Answer - C

Discussion

1305.	Assertion `(A) : ` The presence of `CO_(2)` in the air accelerates corrosion. Reason `(R): CO_(2)` is a poisonous gas.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct.
Answer» Correct Answer - c `(A)` is correct and `(R)` is incorrect. Correct reason `: Fe` in contact with the dissolved `CO_(2)` and `O_(2)` undergoes oxidation faster.

Discussion

1306.	Assertion `(A) :` In a Daniell cell, if the concentration of `Cu^(2+)` and `Zn^(2+)` ions are doubled, the `EMF` of the cell will be doubled. Reason `(R) :` If the concentration of ions in contact with metals is doubled, the electrode potential is doubled.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If `(A)` is incorrect, `(R)` is correct.
Answer» Correct Answer - d

Discussion

1307.	The half cell potentials of a halfcell `A^((x+n)+) , A^(x+)\|pt` were found to be as follows : `{:(% "of reduced form",24.4,48.8),("Half cell potential (V)",0.101,0.115):}` Determinwe the value of `n`.
Answer» Correct Answer - B

Discussion

1308.	Statement-I: `H_(2)+O_(2)` fuel cell gives a constant voltage throughout its life. Because Statement-II: In this fuel cell, `H_(2)` reacts with `OH^(-)` ions yet the overall concentration of `OH^(-)` ions does not change.A. Statement 1 is ture , Statement 2 is true Statement 2 is correct explanation for Statement 10B. Statement 1 is true Statement 2 is ture Statement 2 is NOT a correct explantion for Statement 10C. Statement 1 is true statement 2 is tureD. Statement 1 is false Statement 2 is true
Answer» Correct Answer - A

Discussion

1309.	Explain the terms with suitable exapmples. Molecularity of a reaction
Answer» Molecularity of a reaction : Zero order reaction is the reaction in which rate of reaction does not depends on the concentration of reactants. `R to P` Rate `(-d[R])/(dt)=k [R]^(@)` Rate `=(-d[R])/(dt)` `d[R]=-k.dt` Intergratinng on both sides `[R] =-kt +I-(1)` I = Intergration constant At `t=0 to R=[R]_(0)` initial concentration `I=[R]_(0)` Substituting `I=[R]_(0)` in the above equation (1) `[R] =-kt+[R]_(0)` `k=([R]_(0)-[R])/(t)` This is the integrated rate equation for a zero order reaction.

Discussion

1310.	What is Rate equation (or) Rate expression (or) Rate Law ?
Answer» Rate equation of Rate expression or Rate Law is the methematical expression in which `aA + bB to cC+dD` reaction rate is given in terms of molar concentration of reactants. Where a,b and d are the stoichiometric coeffcients of reactants and products. The rate expression for this reaction is Rate `alpha[A]^(x)[B]^(y)` Exponents x and y may or may not equal to stiochiometric coefficients a and b. Rate `=K [A]^(x)[B]^(y)` Where K is proporationality constant called rate constant.

Discussion

1311.	Define molecularity of a reaction, Illustrate with an example.
Answer» The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction. `NH_(4)NO_(2)to N_(2)+ 2H_(2)O` (Unimolecular) `2HI to H_(2)+I_(2)` (Bimolecular) `2NO+ O_(2)to 2NO_(2)` (Trimolecular)

Discussion

1312.	Why is alternating current used in the measurement of conductivity of the solution ?
Answer» If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.

Discussion

1313.	Define resistivity. What are its units ?
Answer» Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.) It has SI units, ohm m and C.G.S. units, ohm cm.

1313.

Define resistivity. What are its units ?

Answer»

Resistivity (or specific resistance) :

It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, ρ is the resistivity i.e., resistance of a solution of unit volume.) It has SI units, ohm m and C.G.S. units, ohm cm.

Discussion

1314.	Standard electrode potential of An and Fe are known to be (i) -0.76V and (ii) -0.44V respectively. How does it explain that galvanization prevents rusting of iron while zinc slowly dissolves awayA. Since (i) is less than (ii), zinc becomes the cathode and iron the anodeB. Since (i) is less than (ii), zinc becomes the anode and iron the cathodeC. Since (i) is more than (ii), zinc becomes the anode and iron the cathodeD. Since (i) is more than (ii), zinc becomes the cathode and iron the anode
Answer» Correct Answer - B

Discussion

1315.	The potential of the cell `V(s)\|V^(3+)(aq., 0.0011 M) \|\|Ni^(2+)(aq., 0.24 M)\|\|Ni(s)` isA. `0.50 V`B. `0.40 V`C. `0.70 V`D. `0.80 V`
Answer» Correct Answer - C The two half reactions and their standard potential are `{:(V^(3+)(aq.)+3e^(-)hArr V(s)E_(V^(3+)//V)^(@)=-0.89 V),(Ni^(2+)(aq.)+2e^(-) hArr Ni(s) E_(Ni^(2+)//Ni)^(@) = -0.23 V):}` The `V^(3+)//V` couple is the anode thus `E^(@)` for the overall reaction is `E_("cell")^(@) = E_(R)^(@)- E_(L)^(@)` `= (-0.23 V)-(-0.89 V)` `= 0.66 V` To obtain the overlal racito, we reverse the vanadium half-reaction. We multiply it by two and the nikel half-reaction by three.The number of elctrons that cancel when the two half-reactions are combined is six, so `n=6`. The cell reaction is `2V(s)+3Ni^(2+)(aq.)hArr 2V^(3+) (aq.)+3Ni(s)` The expression for `Q` for the overall reaction is `Q = (C_(V3+)^(2))/(C_(Ni2+)^(3))` Using the calcualted value of `E_("cell")^(@)` and the given data on concentration, the potential of the cell can be calculated with the help of the Nernst equation: `E_("cell") = E_("cell")^(@)-(0.0592 V)/(6)log Q` `= (0.66 V)- ((0.0592 V)/(6)) "log" (0.0011)^(2)/(0.24)^(3)` `= 0.70 V` The values of `E_("cell")^(@)` and `E_("cell")` tell us a great deal about the bahaviour of the chemical systeam that corresponds to the cell reaction. The large postivie value of `E_("cell")^(@)` means that the reaction has a large equiblrium constant and proceeds substanitially to completion as written. Since `E_("cell")` is even larger than `E_("cell")^(@)`, the systeam is further from equilibrium than it would be at standard concentrations.

Discussion

1316.	The reducation potential of a `Pt(s)\|Cl_(2)(g)\|Cl^(-)(aq.)` electrode is found to be `1.42 V` when the pressure olf `Cl_(2)` is 0.25 arm. The concentration of chloride ion in this half cell isA. `0.43 M`B. `0.043 M`C. `0.34 M`D. `0.034 M`
Answer» Correct Answer - B The half cell reaction is `Cl_(2)(g) + 2e^(-) hArr 2Cl^(-)(aq.)` and its `E_(Cl_(2)//Cl^(-)` is `1.36 V` The value of `n`, the number of elctrons transferred, is `2`. The expression for `Q` is `C_(Cl^(-)//P_(Cl_(2)))` Substituting these data into the Nernst equation gives `E = E^(@) -(0.0592 v)/(n) log Q` `1.42 V = 1.36 V - (0.0592 V)/(n) "log" (C_(Cl^(-))^(2))/(0.25)` `C_(Cl^(-)) = 0.043 M`

Discussion

1317.	Fluorine is the best oxidising agent because it hasA. highest electron afffinityB. highest reduction potentialC. highest oxidation potentialD. lowest electron affinity.
Answer» Correct Answer - B Higher the reduction potential, stronger is the oxidising agent

Discussion

1318.	The observed emf of the cell, `Pt\|H_(2) ("1 atm")\|H^(+)(3xx10^(-4) M)\|\|H^(+) (M_(1))\|H_(2) ("1 atm")\|Pt` is 0.154 V. Calculate the value of `M_(1)` and pH of cathodic solution.
Answer» `E_(cell)=0.0591" log" M_(1)/(3xx10^(-4))` or `"log"M_(1)/(3xx10^(-4))=0.154/0.0591=2.6058` `M_(1)/(3xx10^(-4))=4.034xx10^(2)` `M_(1)=4.034xx10^(2)xx3xx10^(-4)M` `=0.121 M` `pH=-log[H^(+)]=-log 0.121=0.917`

Discussion

1319.	Ionic mobility`=("Ionic conductance")/("____")`
Answer» Correct Answer - 96500

Discussion

1320.	Which of the following shows electrical conduction ?A. PotassiumB. GraphiteC. DiamondD. Sodium
Answer» Correct Answer - B Graphite is a good conductor of electricity .

Discussion

1321.	The unit for the electric current isA. OhmB. VoltC. AmpereD. Coulomb
Answer» Correct Answer - C

Discussion

1322.	The quantitiy of electricity required to liberate 112 `cm^(3)` of hydrogen at STP from acidified water isA. 0.1 FaradayB. 1 faradayC. 965 coulombD. 96500 coulomb
Answer» Correct Answer - C

Discussion

1323.	The quantitiy of electricity required to liberate 112 `cm^(3)` of hydrogen at STP from acidified water isA. `965C`B. 1FaradayC. 0.1FD. 96500C.
Answer» Correct Answer - A `2H^(+)+underset(2F)(2e^(-))rarrunderset(22400cm^(3))(H_(2))` To liberate 22400 `cm^(3)` at STP electricity required =`2xx96500C` To liberate 112 `cm^(2)` at STP electricity required `=(2xx96500xx112)/(22400)` =965C

Discussion

1324.	Which of the following methods does not liberate hydrogen ?A. `Zn+H_(2)SO_(4)` (dil.)B. `Mg+H_(2)SO_(4)`(dil.)C. `Cu+H_(2)SO_(4)`(dil.)D. `Zn+HCl` (dil.)
Answer» Correct Answer - C

Discussion

1325.	A depolarizer used in dry cell is :A. ammonium chlorideB. manganese dioxideC. potassium oxideD. sodium phosphate
Answer» Correct Answer - B

Discussion

1326.	The element which can displace three other halogens from their compound is :A. FB. ClC. BrD. I
Answer» Correct Answer - A

Discussion

1327.	Assertion Zinc and iron decompose steam whereas copper and mercury do not. Reason A metal can displace hydrogen from water only if its reduction potential is less than that of hydrogen.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false.
Answer» Correct Answer - A

Discussion

1328.	The element which can be displace three other halogens from their compound is:-A. ClB. FeC. BrD. I
Answer» Correct Answer - B Because flourine is most powerful reducing agent than other halogens.

Discussion

1329.	I the electrochemical cell `H_(2(g))1atm\|H^(+)(1M)\|\|Cu^(2+)(1M)\|Cu_((s))` Which one of the following statements is true:-A. `H_(2)` is cathode, Cu is anodeB. Oxidation occurs at Cu electrodeC. Reduction occurs at `H_(2)` electrodeD. `H_(2)` is anode, Cu is cathode
Answer» Correct Answer - D `H_(2)` is anode because oxidation takes place. Cu is cathode because reduction takes place.

Discussion

1330.	EMF of a cell whose half cells are given below is `Mg^(2+)+2e^(-)toMg(s),E=-2.37V` `Cu^(2+)+2e^(-)toCu(s),E=+0.33V`
Answer» Correct Answer - C `E_(cell)^(o)=E_("cathode")^(o)-E_("anode"), " "E_(cell)^(o)=0.34-(-2.37)` `E_(cell)^(o)=2.71V`

Discussion

1331.	The volume of `H_(2)` gas at NTP obtained by passing 4 amperes through acidified `H_(2)O` for 30 minutes isA. 0.836LB. 0.0432LC. 0.1672LD. 5.6L
Answer» Correct Answer - A Quantity of electricity passed `=4Axx(30xx60s)=7200C` `2H_(2)O+2e^(-)rarrH_(2)+20H^(-)` `2xx96500C` liberate `H_(2)=22.4L` at STP 7200C will liberate `=(22.4)/(2xx96500)xx7200C` `=0.836L`

Discussion

1332.	Which of the following expressions is no correct ?A. `mu^(oo) = gamma_(+)lambda_(+)^(oo) + gamma_(-)lambda_(-)^(oo)`B. `lambda^(oo) = (1)/(n+)lambda_(+)^(oo)+(1)/(n-)lambda_(+)^(oo)`C. `lambda_("cation")^(oo) = u_("cation")^(oo) xx "faraday"`D. `lambda_(anion)^(oo) = u_(anion)^(oo) xx faraday`
Answer» Correct Answer - D `lambda_("anion")^(oo) = u_("anion")^(oo) xx "faraday"`

Discussion

1333.	The equivalent conductance of `1 M` benzoic acid is `12.8 ohm^(-1) cm^(2)`. If the conductance of benzoate ion and `H^(+)` ion are `12` and `288.42 ohm^(-1) cm^(2)` respectively. Its degree of dissociation is :A. `39 %`B. `3.9 %`C. `0.35 %`D. `0.039 %`
Answer» Correct Answer - B `overset(@)(Lambda)(C_(6)H_(5)COOH) = overset(@)(Lambda)(C_(6)H_(5)COO^(-)) + overset(@)(Lambda)_((H^(+)))` `= 42 xx 288.42 = 330.42` `alpha = (Lambda_(m)^(C))/(overset(@)(Lambda)_(m)) = (12.8)/(330.42) = 0.0387 = 3.87%`

Discussion

1334.	Why do platinum and gold occur in free state in nature ?
Answer» Platinum and gold have high negative values of oxidation potentials. They do not possess tendency to form positive ions at all. Therefore, they do not occur in combined state unlike other metals and hence, they occur free in nature.

Discussion

1335.	The amount of lactic acid, `HC_(3)H_(5)O_(3)`, produced in a sample of muscle tissue was analysed by reaction with hydroxide ion. Hydroxide ion was produced in the sample mixture by electrolysis. The cathode reaction was, `2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-) (aq.)` Hydroxide ion reacts with lactic acid as soon as it is produced. The end point of the reaction is detected with an acid-base indicator. It required 115 seconds for a current of 15.6 mA to reach end point. How many grams of lactic acid (a monoprotic acid) were present in the sample ?
Answer» Correct Answer - `1.674 xx 10^(-3) g` No. of moles of lactic acid = No. of moles of `OH^(-)` used =No. of faraday used in electrolysis Number of faradat used `=(I xx t)/(96500) =(15.6xx10^(-3) xx115)/(96500)` `= 1.86 xx 10^(-5)` Mass of lactic acid = Number of moles `xx` Molecular mass `=1.86 xx 10^(-5) xx 90 =1.674 xx 10^(-3) g`

Discussion

1336.	How many faradays of eletricity is required to deoposit 2 mot copper from `CuSO_(4)` solution
Answer» Correct Answer - D

Discussion

1337.	Consult the table on standard electrode potentials and suggest three substance that can oxidise ferrous ions under suitable conditions.
Answer» `Fe^(2+)` (ferrous) ions gets oxidised to `Fe^(3+)` (farriec) ion as follows `Fe^(2+) to Fe^(3+)+e^(-)E_(0x)^(0)=-0.77V` Only those substances can oxidise `Fe^(2+)` ions to `Fe^(3+)` ions which can accept electrons released during oxidation or are placed above iron in electrochemical series. Three such substances are `Cl_(2)(g), Br_(2)(g) and F_(2)(g).`

Discussion

1338.	Consult the table of the standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.
Answer» For the oxidation of ferrous ion, following reaction will take place Fe²⁺ → Fe³⁺ + e^– E°_ox = – 0.77 V Only those substances can oxidise ferrous ion into ferric ion, which are stronger oxidising agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reeaction is positive. This is so for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series e.g Br₂, Cl₂, and F₂.

Discussion

1339.	Calculate the degree of dissociation `(alpha)` of acetic acid if its molar conductivity is 39.05 S `cm^(2)mol^(-1)`. Given `lamda^(@)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(@)(CH_(3)CO O^(-))=40.9" S "cm^(2)mol^(-1)`.
Answer» `alpha=(wedge_(m))/(wedge_(m)^(@))` `wedge_(m)^(@)(CH_(3)COOH)=lamda^(@)(CH_(3)COO^(-))+lamda^(@)(H^(+))=40.9+349.6=390.5" S "cm^(2)mol^(-1)` `wedge_(m)=39.05" S "cm^(2)mol^(-1)` `thereforealpha=(39.05)/(390.5)=0.1`

Discussion

1340.	The e.mf. of an electrolytic cell depends mostly on :A. The valency of the reacting meterialsB. The quantity of electrolytic solutionC. The size of the electrodesD. The density of the electrolytic solution
Answer» Correct Answer - A `E_("cell") = E^(@) -(2.303RT)/(n) log[("Oxi.")/("Red")]` `= E^(@) -(0.059)/(n) log[("Oxi.")/("Red")]` at `25^(@)C` cosntant temperature mostly `E_("cell")` depends on the `n` wherer is the valency of the reacting materials.

Discussion

1341.	A solution containing one mole per litre each of `AX, BX_(2), CX_(2)` and `DX_(2)` is electrolysed using inert electrodes. The values of the standard potentials for reduction reactions of `A^(+)\|A, B^(2+)\|B,C^(2+)\|C` and `D^(2+)\|D` are `+.80, +.34, -0.76` and `-1.66` volts respectively. The correct sequence in which these metals will be deposited on the cathode is :A. `A, B, C, D`B. `D, C, B, A`C. `A, C, B, D`D. `D, B, C, A`
Answer» Correct Answer - A The metal ion which is discharged first has higher reduction potential i.e. , `A gt B gt C gt D`

Discussion

1342.	For which of the following compounds are the products of the electrolysis (using inert electrodes) of a fused metal salt and a concentrated aqueous solution most likely to be the same ?A. Copper `(II)` chlorideB. Magneisum bromideC. Potassium iodideD. Sodium hydroxide
Answer» Correct Answer - A Anode : `2Cl^(-) rarrCl_(2) + 2e` Cathode : `Cu^(2+) + 2e rarr Cu`

Discussion

1343.	Which one of the following material conducts electricityA. DiamondB. Crystalline sodium chlorideC. Barium sulphateD. Molten sulphur
Answer» Correct Answer - D Generally fused potassium chloride show the electric.

Discussion

1344.	A silver electrode is immersed in saturated `Ag_(2)SO_(4(aq))`. The potential difference between the silver and the standard hydrogen electrode is found to be `0.711V` Determine `K_(SP)(AgSO_(4))`. Given `E_(Ag^(+)//Ag)^(@)=0799V`.
Answer» The given cell is `PtH_(2)\|(H^(+)),(1M)\|\|(Ag_(2)SO_(4(aq.))),("saturated")\|Ag` The reaction are, `H_(2) rarr 2H^(+) + 2e` `2Ag^(+) + 2e rarr 2Ag` Thus, `E_(cell) = E_(OP_(H)) + E_(RP_(H))` `= 0.711 = 0.799 + (0.059)/(2)log[Ag^(+)]^(2)` `:. log[(1)/([Ag^(+)]^(2))] = ([0.799 - 0.711] xx 2)/(0.059) = 3` `:. [Ag^(+)]^(2) = 10^(-3)` `:. [Ag^(+)]^(2) = 3.2 xx 10^(-2)` Now the solubility equilibrium is, `Ag_(2)SO_(4) hArr 2Ag^(+) + SO_(4)^(2-)` `K_(SP) = (Ag^(+))^(2)(SO_(4)^(2-))` `= (3.2 xx 10^(-2))^(2)((3.2 xx 10^(-2))/(2))` `= 1.6 xx 10^(-5)` (Note that if `[Ag^(+)] = 3.2 xx 10^(-2)` then `[SO_(4)^(2-)] = (1)/(2) xx 3.2 xx 10^(-2))`

Discussion

1345.	Find the solubility product of a saturated solution of `Ag_(2)CrO_(4)` in water at `298K`, if the `EMF` of the cell `:` `Ag\|Ag^(o+)(satAg_(2)CrO_(4)sol)\|\|Ag(0.1M)\|Agis 0.164V` at `298K`.
Answer» For the cell `Ag\|Ag^(+)(Ag_(2)CrO_(4) "sol. Saturated") \|\|(Ag^(+)),(0.1M)\|Ag`: `E_(cell) = 0.164 V` at `298 K` We have `E_(cell) = E_(OP_(Ag//Ag^(+)))^(@)+E_(RP_(Ag^(+)//Ag))^(@)` `+ (0.059)/(1)log_(10).([Ag^(+)]_(R.H.S.))/([Ag^(+)]_(L.H.S.))` or ` 0.164 = 0 + (0.059)/(1)log_(10).(0.1)/([Ag^(+)]_(L.H.S.))` `:. [Ag^(+)]_(L.H.S.) = 1.66 xx 10^(-4)M` Now `K_(SP)` for `Ag_(2)CrO hArr 2Ag^(+) +CrO_(4)^(2-)` `K_(SP) = [Ag^(+)]^(2)[CrO_(4)^(2-)]` Since, `[Ag^(+)]_(L.H.S.) = 1.66 xx 10^(-4)M` `:. [CrO_(4)^(2-)]_(L.H.S.) = (1.66 xx 10^(-4))/(2) M` `:. K_(SP) = [1.6 xx 10^(-4)]^(2)[(1.66 xx 10^(-4))/(2)]` `K_(SP) = 2.287 xx 10^(-12) mol^(3) litre^(-3)`

Discussion

1346.	Find the solution product of `Ag_(2)CrO_(4)` in water at 298 K if the e.m.f. of the cell `Ag//Ag^(+)` (saturated `Ag_(2)CrO_(4)soln.)\|\|Ag^(+)(0.1M)//Ag" is "0.164V" at "298K`.
Answer» The cell is `Ag\|Ag^(+)(c_(1))\|\|Ag^(+)(0.1M)\|Ag` `E_(cell)=(0.0591)/(n)"log"(c_(2))/(c_(1))` (for concentration cell) `0.164=(0.0591)/(1)"log"(0.1)/(c_(1))` or log`(0.1)/(c_(1))=2.7750` or `0.1//c_(1)=5.957xx10^(2)` o `c_(1)=1.679xx10^(-4)` i.e., `[Ag^(+)]` in saturated `Ag_(2)CrO_(4)` sol.=`1.679xx10^(-4)M` `Ag_(2)CrO_(4)hArr2Ag^(+)+CrO_(4)^(2-)` `[Ag^(+)]=1.679xx10^(-4)M,[CrO_(4)^(2-)]=([Ag^(+)])/(2)=(1.679xx10^(-4))/(2)M` `K_(sp)=[Ag^(+)]^(2)[CrO_(4)^(2-)]=(1.679xx10^(-4))^(2)xx((1.679xx10^(-4))/(2))=2.37xx10^(-12)`.

Discussion

1347.	Which metal can deposit copper from copper sulphate solution ?A. mercuryB. ironC. goldD. platinum
Answer» Correct Answer - B Only iron is placed below Cu in electrochemical series (CAHPA) .

Discussion

1348.	Which of the following metal can deposit copper from copper sulphate solutionA. MercuryB. IronC. GaldD. Platinum
Answer» Correct Answer - B `Cu^(++)` will be reduced and Fe will be oxidized.

Discussion

1349.	What is the potential of the cell containing two hydrogen electrode as represented below ? `Pt,(1)/(2) H_(2)(g)\|H_(2)O\|\|H^(o+)(0.01M)\|1//2H_(2)(g)Pt`A. `-0.236V`B. `-0.0591V`C. `0.236V`D. `0.0591V`
Answer» Correct Answer - c `E_(cell)=(0.059)/(1)log.([H^(o+)]_(RHS))/([H^(o+)]_(LHS))` `[` For `H_(2)O,[H^(o+)]=[overset(c-)(O)H]=10^(7)M]` `=0.059log.(10^(-3))/(10^(-7))=0.059xx4=0.236V` or `E_(cell)=-0.059(pH_(c)-pH_(a))=-0.059(3-7)=0.235V`

Discussion

1350.	How long will it take for a uniform current of 6.0 ampere to deposite 78.0 g gold from asolution of `AuCl_(4)^(-)` ? What mass of chlorine gas will be formed simultaneously at the anode in the electrolytic cell ?
Answer» Correct Answer - 19104 second; chlorine liberated = 42.17 g `AuCl_(4)^(-)+underset(3F)(3e^(-)) rarr underset("1 mole")(Au)+4Cl^(-) ("Cathode")` `2Cl^(-) rarr underset("1 mole")(Cl_(2))+ underset(2 F)(2e^(-)) ("Anode")`

Discussion

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