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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1151. |
In ` H_2 -O_2` fuel cell, `6.72 L` of hydrogen at `NTP` reacts in `15` minutes, the averge current produced in ampres is .A. ` 64.3` ampB. ` 643 .3` ampC. ` 6. 43` ampD. `0.643` amp |
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Answer» Correct Answer - A ` (i xx 15xx 60)/( 96500) = (6.72)/(22.4) xx 2 rArr i=64.3 A`. |
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| 1152. |
In a hydrogen oxyge fuel cell, electricity is produced. In this process `H_2`(g) is oxided at anode and `O_2`(g) reduced at cathode Given: Cathode `O_2(g)+2H_2O(l)+4e^(-)to4OH^(-)(aq)` Anode `H_2(g)+2OH^(-)(aq)to2H_2O(l)+2e^(-)` 4.48 litre `H_2` at 1atm and 273 k oxidised in 9650 sec. If current produced in fuel cell, is used for the deposition of `Cu^(+2)` in 1L,2M `CuSO_4`(aq) solution for 241.25 sec using Pt. electrode, the pH of solution after electrolysis is:A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B |
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| 1153. |
10g fairly concentrated solution of `CuSO_(4)` is electrolyzed using `0.01F` of electricity. Calculate: (a) The weight of resulting solution (b) Equivalents of acid or alkali in the solution. |
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Answer» Electrode process during electrolysis of aqueous `CuSO_(4)` may be given as: `Cu^(2+) + 2e^(-) rarr Cu` (Cathode) `2OH^(-) rarr H_(2)O + 1//2 O_(2) + 2e^(-)` (Anode) Mass of copper deposited at cathode by 0.01 faraday charge `= 0.01 xx 31.75` `= 0.3175g` (Here, 31.75 is the equivalent mass of `Cu^(2+)`) Mass of oxygen evolved by 0.01 faraday charge `= 0.01 xx 8 = 0.08 g` Total weight loss from solution `= 0.3175 + 0.08 = 0.3975 g` Mass of resulting solution `= 10 - 0.3975` `= 9.6025 g` After deposition of `Cu^(2+) and OH^(-)` ions at the respective electrodes, `H_(2)SO_(4)` will prevail in the solution. 0.01 faraday of electricity will result in 0.01 equivalent of acid. |
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| 1154. |
A constant current was flowen for `1 mi n` through a solution of `Kl` . At the end of experiment, liberated `I_(2)` consumed `150mL` of `0.01M` solution of `Na_(2)S_(2)O_(3)` following the reaction `:` `I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(c-)+S_(4)O_(6)^(2-)` What was the average rate of current flow in ampere ? |
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Answer» `2S_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-)+2e^(-)" "(n` factor `=(2)/(2)=1)` `mEq` of `Kl=mEq` of `I_(2)` liberated `=mEq` of `Na_(2)SO_(3)` `=150 mLxx0.01xx1(n` factor `)` `=1.5mEq` `[1F=96500C=1 Eq` or `S_(2)O_(3)^(2-)]` `1.5xx10^(-3)Eq=96500Cxx1.5xx10^(-3)Eq` `implies I xx t(s)=Ixx60` `:. I=(96500Cxx1.5xx10^(-3)Eq)/(60s)=(965)/(400)=2.4125A` |
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| 1155. |
A current of `1.70A` is passed trhough `300.0 mL` of `0.160M` solution of `ZnSO_(4)` for `230s` with a current efficiency of `90%`. Find out the molarity of `Zn^(2+)` after the deposition of `Zn`. Assume the volume of the solution to remain constant during the electrolysis. |
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Answer» Correct Answer - `0.154M` `I=(1.70xx90)/(100)A` `:.` Equivalent of `Zn^(2+)` lost `=(It)/(96500)` `=(1.70xx90xx230)/(100xx96500)` `=3.646xx10^(-13)` Milliequivalent of `Zn^(2+)` lost `=3.646xx10^(-3)` Initially milliequivalent of `Zn^(2+)=300xx0.160xx2=96` `( :. Mxx2=N fo r Zn^(2+)impliesmEq=NxxV_((mL)))` `:.` Milliequivalent of `Zn^(2+)` left in solution `=96-3.646` `=92.354` `:. [ZnSO_(4)]=(92.254)/(2xx300)=0.154M` |
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| 1156. |
A current of 1.70 A is passed through 300.0 mL of 0.160 M solution of a `ZnSO_(4)` for 230 s with a current efficiency of 90% . Find out the molarity of `Zn^(2+)` after the deposition Zn. Assume the volume of the solution to remain cosntant during the electrolysis. |
| Answer» Correct Answer - 0.154M | |
| 1157. |
For cell reaction, `Zn+Cu^(2+)toZn^(2+)+Cu`, cell representation isA. `Zn|Zn^(2+)||Cu^(2+)|Cu`B. `Cu|Cu^(2+)||Zn^(2+)|Zn`C. `Cu|Zn^(2+)||Zn|Cu^(+)`D. `Cu^(2+)|Zn||Zn^(2+)|Cu` |
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Answer» Correct Answer - A Electrode on which oxidation occurs is written on L.H.S. and the other on the R.H.S. as represented by `Zn|Zn^(2+)||Cu^(2+)|Cu`. |
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| 1158. |
During the electrolysis of a concentrated brine solution, Calculated the moles of chlorine gas produced by the passage of 4F electricity. |
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Answer» Correct Answer - B |
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| 1159. |
Assertion:In mercury cell, the cell potential is approximately 1.35 V Which cell will measure standard electrode and remains constant during its life. Reason : The overall reaction in mercury cell is represented as `Zn(Hg)+HgO_((s)) to ZnO_((s)) + Hg_((l))`A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - A Overall reaction does not involve any ion in solution whose concentration can change during its life time . |
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| 1160. |
Strong electrolytes are those which:A. conduct electricityB. dissolve readily in waterC. dissolve into ions at high dilutionD. completely dissociation into ions |
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Answer» Correct Answer - D |
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| 1161. |
Assertion:In electrolysis of aqueous NaCl the product obtained is `H_2` gas. Reason:Gases are liberted faster than the metals.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - C At cathode there is competition between the following reduction reactions : `Na_((aq))^(+)+e^(-) to Na_((s)) , E_("cell")^@=-2.71 V` `H_((aq))^(+) + e^(-) to 1/2H_(2(g)) , E_"cell"^@`=0.00 V Hence, reaction with higher `E^@` value is preferred and `H_2` is liberated. |
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| 1162. |
Brine solution on electrolysis will not give.A. NaOHB. `Cl_2`C. `H_2`D. `O_2` |
| Answer» Correct Answer - D | |
| 1163. |
Strong electrolytes are those which:A. Dissolve readily in waterB. Conduct electricityC. Dissociate into ions at high dilutionD. Completely dissociate into high ions at all dilutions |
| Answer» Correct Answer - D | |
| 1164. |
Which of the following metals will give `H_2` on reaction with `NaOH`.A. MgB. BaC. CaD. Sr |
| Answer» Correct Answer - A | |
| 1165. |
The ionic conductance of the following cations in a given concentration is in the orderA. `Li^(+)ltNa^(+)ltK^(+)ltRb^(+)`B. `Li^(+)gtNa^(+)gtK^(+)gtRb^(+)`C. `Li^(+)ltNa^(+)gtK^(+)gtRb^(+)`D. `Li^(+)=Na^(+)ltK^(+)ltRb^(+)` |
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Answer» Correct Answer - A Due to extensive hydration of smaller alkali metal cations, the ionic radii is in the order `Li^(+)gtNa^(+)gtK^(+)gtRb^(+)` As such mobility in water is in the order `Li^(+)ltNa^(+)ltK^(+)ltRb^(+)` |
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| 1166. |
If the current is passed into the solution of an electrolyteA. Anions move towards anode, cations towards cathodeB. Anions and cations both move towards anodeC. Anions move towards cathode, cations towards anodeD. No movement of ions take place |
| Answer» Correct Answer - A | |
| 1167. |
Which solution will show highest resistance during the passage of currentA. 0.05 N NaClB. 2N NaClC. 0.1 N NaClD. 1N NaCl |
| Answer» Correct Answer - B | |
| 1168. |
The value of one Faraday isA. `95500C mol^-1`B. `96550 C mol^-1`C. `96500 Cmol^-1`D. `98500 C mol^-1` |
| Answer» Correct Answer - C | |
| 1169. |
An electric c urrent is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108g more at the end of the electrolysis. The volume of oxygen evolved at STP:A. `56cm^(3)`B. `550cm^(3)`C. `5.6cm^(3)`D. `11.2cm^(3)` |
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Answer» Correct Answer - C `("Mass of Ag deposited")/("Mass of" O_(2) "evolved")=("Eq. mass of Ag")/("Eq. mass of" O_(2))` `(0.108)/(m(O_(2)))=(108)/(8)` `m(O_(2))=(8xx0.108)/(108)=8xx10^(-3)g` `32gO_(2)=22400 cm^(3)` at N.T.P. `therefore 8xx10^(-3)g "of" O_(2)=(22400)/(32)xx8xx10^(-3)cm^(3)` at N.T.P. `=5.6cm^(3)` at N.T.P. |
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| 1170. |
An electric current is passed through silver voltameter connected to a water voltmeter. The cathode of the silver voltameter is 0.108g more at the end of the electrolysis. The volume of oxygen evolved at STP:A. `56cm^(3)`B. `550cm^(3)`C. `5.6cm^(3)`D. `11.2cm^(3)` |
| Answer» Correct Answer - C | |
| 1171. |
Which of the following isnot a strong electrolyte?A. NaClB. `KNO_(3)`C. `NH_(4)OH`D. `FeSO_(4)` |
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Answer» Correct Answer - C `NH_(4)OH` is a weak electrolye. |
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| 1172. |
The quantity of electricity required to deposit 54 g of silver from silver nitrate solution is : (a) 0.5 Coulomb (b) 0.5 Ampere (c) 0.5 Faraday (d) 0.5 Volt |
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Answer» Option : (c) 0.5 Faraday |
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| 1173. |
The conductivity of a strong electrolyte:A. increases on dilution slightlyB. does not change on dilutionC. decreases on dilutonD. depends on density of electrolyte itself. |
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Answer» Correct Answer - A See comprehensive review for detials. |
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| 1174. |
The standard emf of the cell `Zn+Cu^(2+)rarrCu+Zn^(2+)` is 1.10V at `25^(@)c` the emf of the cell when 0.1 M `Cu^(2)+` and 0.1 M `Zn^(2+)` solution are used will beA. 1.10VB. 0.110VC. `-1.10V`D. `-0.1 10V` |
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Answer» Correct Answer - A `E_("cell")=E_("cell")^(@)-(0.059)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` `=1.10-(0.059)/(2)"log"(0.1)/(0.1)` =1.10V |
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| 1175. |
Which of the following undergoes Hydrolysis most easily:A. B. C. D. |
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Answer» Correct Answer - D If there is more m-directing group then there will be more nuclephilic substitution reaction. |
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| 1176. |
The product in the following reaction is: `Ph-CI +Fe//Br_(2)rarr` ProductA. o-bromo-chloro benzeneB. p-bromo-chloro benzeneC. A and B bothD. 2,4,6-tribromo chloro benzene |
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Answer» Correct Answer - C Since-`CI` group is deactivating and `o//p` directing group so only o-and p-products are formed. |
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| 1177. |
The most reactive towards `SN^(1)` is:A. `PhCH_(2)CI`B. `Ph-CI`C. `CH_(3)CHCI(CH_(3))`D. `p-NO_(2)-Ph-CH_(2)-CI` |
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Answer» Correct Answer - A `S_(N^(1))` the intermediate carbocation is formed. `C_(6)H_(5)-CH_(2)CI rarr C_(6)H_(5) overset(o+)(CH_(2))` is maximum stable due to resonance. |
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| 1178. |
Which of the following plots represents correctly the variation of conductivity with concentration?A. B. C. D. |
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Answer» Correct Answer - C This is the general shape of the a conductivity-con-centration curve for both weak and stron electrolytes. The curves for both types of electrolytes show oa maximum, is only obtained when a wide range of concentrations is possible. In many cases, a sturateed solution is obtained before the maximum is reached so that only the left hand part of the curve is obtainable. As a solution gets more and more concentrated there are more and more solute particles in a given volume of the solution. Because it is the solute that causes electrical conductivity [pure water and other solvents are very poor conductors) it might on this account, be expected, that conductivity and, at first, for dilute soltutions, it does so. However, at higher concentrations the conductivity reaches a maximum and then begins to fall, and some factor other than the increase in the number of solute particles on increasing concentration must enter in. For a weak electrolyte, this factor is the degree of ionization which decreases as the concentration increases. The maximum in the curve is due to the composite effect of an increases in the total number of ionization. For a strong electrolytes, the other factor is the increase in ionic interference (decrease in ionic freedom) as the concentration increases. A strong electrolyte is fully, or almost fully, inoized at all concentrations. However, at high concentrations the inos will interface with each other so that their freedom of movement and their speed are restricted. As a solution gets more concentrated, the ionic interference increases and the ionic speed decreases. Thus, for a strong electrolyte, increasing concentration gived more ions, in a fixed volume of solution, but the ions that are present inter-fere with each other, which gives lower ionic speeds. |
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| 1179. |
The `EMF` of the cell, `Cr|Cr^(+3) (0.1M) ||Fe^(+2) (0.01M)|Fe` (Given: `E^(0) Cr^(+3)|Cr =- 0.75 V, E^(0) Fe^(+2)|Fe =- 0.45 V)` |
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Answer» Half cell reactions are: `{:("At anode",:[Crrarr Cr^(+3) +3e^(-)]xx2),("At anode":,underline([Fe^(+2)+2e^(-)rarrFe]xx3)),("Over all reaction":,2Cr+3F^(+2)rarr2Cr^(+3)+3Fe):}` `E_(cell)^(0) =` Oxidation pot.`+` Reduction pot. `=0.75 +(-0.45) = 0.30` `E_(cell) = E^(0) - (0.0591)/(n) log .(["Product"])/(["Reactant"])` `= 0.30 - (0.0591)/(6) log .([Cr^(+3)]^(2))/([Fe^(+2)]^(3))` `= 0.30 - (0.0591)/(6) log .[0.1]^(2)/([0.01]^(3))` `= 0.30 - (0.24)/(6) =0.26` volt. |
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| 1180. |
Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at `25^(@)C`. The standard electrode potential of `Cu^(2+)//Cu` system is 0.34 volt at 298 K. |
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Answer» We know that, `E_("red")=E_("red")^(@)+0.0591/n log_(10) ["ion"]` Putting the value of `E_("red")^(@)=0.34 V, n=2` and `[Cu^(2+)]=0.1 M` `E_("red")=0.34+0.0591/2 log_(10) [0.1]` `=0.34+0.02955xx(-1)` `=0.34-0.02955=0.31045` volt |
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| 1181. |
Which of the following increases with dilution?A. ConductivityB. Molar conductivityC. Equivalent conductivitiyD. both (2) and (3) |
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Answer» Correct Answer - D Both molar and equivalent conductivities increases with decreases in concentration because the total volume, `V`, of solution containing one mole (or one equivalent) of the electrolyte increases more than decrease in conductivity. The increase in the tendency to form ion pairs while the increase in the case of weak electrolytes is due to increases in degree of ionization |
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| 1182. |
The same current if passed through solution of silver nitrate and cupric salt connected in series. If the weights of silver deposited is `1.08g`. Calculate the weight of copper deposited |
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Answer» According to faradays second law. `(W_(1))/(W_(2)) = (E_(1))/(E_(2)) rArr (1.08)/(W_(2)) = (108)/(31.75) rArr W_(2) = 0.3175 g` |
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| 1183. |
The `0.1M` copper sulphate solution in which copper electrode is dipped at `25^(@)C`. Calculate the electrode potential of copper electrode. [Given : `E^(0) Cu^(+2)//Cu = 0.34V]` |
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Answer» `Cu^(+2) +2e^(-) rarr Cu` `E_(red) = E_(red)^(0) - (0.0591)/(n) log .(["Product"])/(["Reactant"])` Here `n = 2` So `E = 0.34 -(0.0591)/(2) log 10` `= 0.34 - 0.03 = 0.31` volts |
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| 1184. |
Which of the following is correct regarding the variation of conductivity with dilution ?A. It decreases for strong electrolytic but increases weak electroyltesB. It decreases for both strong and weak electrolytesC. It increases for both strong electrolytes and weak electrolnytesD. It increases for strong electrolytes but decreases weak electrolytes |
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Answer» Correct Answer - B Conductivity always decreases with decreases in con-centration for weak as well as strong electrolytes. This can be explanied by the fact that the number ions per unit volume that carry the current in a solution decreases on dilution |
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| 1185. |
A cell constant is generally found by measuring the conductivity of aqueous solution ofA. `BaCl_(2)`B. `KCl`C. `NaCl`D. `MgCl_(2)` |
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Answer» Correct Answer - b `KCl` is used since its conductivity is known accurately at various concentrations and at different temperatures. |
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| 1186. |
The cell constant of a conductivity cell is usually determined by measuring the resistance of the elctrolytic solutions `"_____"` whose conductivity is already known accurateble various concentrations and at different temperatures.A. `KCl`B. `HCl`C. `NaCl`D. `LiCl` |
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Answer» Correct Answer - A The quantity `1//A` is called cell constant denoted by the symbol `G^(**)`. It depends on the distance between the electrodes and their area of cross-section and has the dimension of `"length"^(-1)` and can be calculated if we knoew `l` and `A`. However, measurement of `l` and `A` is not only inconlvenient but also unreliable . `KCl` solutions are generally used because their conductivity is known accurately at various concentrations and temperatures. |
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| 1187. |
A cell constant is generally found b y mesuring the conductivity of aqueous solution ofA. `BaCl_(2)`B. `KCl`C. `NaCl`D. `MgCl_(2)`. |
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Answer» Correct Answer - B Conceptual question |
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| 1188. |
When during electrolusis of a solution of `AgNO_3, 9650` coulmbs of charge pass through the electroplationg bath, the mass of silver deposited on the cathode will be:A. 1.08gB. 10.8gC. 21.6gD. 108g |
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Answer» Correct Answer - B `Ag^(+) overset(+e^(-))toAg`, 96500 C will liberate silver=108 gm. 9650C will liberate silver=10.8gm. |
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| 1189. |
`I_(2)` and ` F_(2)` are added to a solution containing 1M each of `I^(-) and F^(-)`. What reaction will take place? Given that the reduction potential of `I_(2) and F_(2)` are 0.54 volt and 2.87 volts respectively. |
| Answer» `F_(2)+2I^(-)to2F^(-)+I_(2)` because for this reaction EMF will be +ve. | |
| 1190. |
An electrochemical cell is made of aluminium and tin electrodes with their standard reduction potentials -1.66 V and 0.14 V respectively. Select the anode and the cathode, represent the cell and write the cell reaction. Find the e.m.f. of the cell |
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Answer» For `E_(cell)` to be positive, the given standard reduction potentials show that oxidation will take place on Al-electrode. Hence, Al will be anode and Sn will be cathode. The cell will be represented as: `Al(s)|Al^(3+)(aq)||Sn^(2+)(aq)|Sn(s)` ltBrgt `E_(cell)^(@)=E_(Sn^(2+),Sn)^(@)-E_(Al^(3+),Al)^(@)=0.14-(-1.66)=1.80V` |
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| 1191. |
Consider the statement ‘Specific conductance increases on dilution’.1. Is it true? 2. What is your opinion about it? Justify. 3. Dene specific conductance and give its equation. |
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Answer» 1. No. 2. This is because, the number of ions per unit volume that carry the current in a solution decreases on dilution. 3. Specific conductance/Conductivity of a conductor is the conductance offered by the conductor of unit length and unit area of cross section. k = \(\frac{l}{RA}=\frac{1}{R}\,\times\,\frac{l}{A}\) is the cell constant. |
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| 1192. |
“Metal having -ve reduction potential when dipped in the solution of its own ion has a tendency to go into the solution”.1. Analyse the above statement.2. Compare the reduction potential of H+ ion with that of other metal ions. |
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Answer» 1. The statement is correct. 2. The reduction electrode potentials of elements lying above hydrogen in the electrochemical series are negative while those of elements lying below hydrogen are positive. The standard potential for the reaction 2H+ + 2e- → H2(g) is taken as 0.00 V. |
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| 1193. |
Standard reduction potential of an element equal toA. `+ 1 xx` its reduction potentialB. `- 1 xx ` its standard oxidation potentialC. `0.00`VD. `+ 1 xx ` its standard oxidation potential |
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Answer» Correct Answer - B `E_(OP)^(@) = - E_(RP)^(@)` for any element . |
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| 1194. |
A metal having negative reduction potential when dipped in the solution of its own ions, has a tendencyA. to pass into the solutionB. to be deposited from the solutionC. to become electrically positiveD. to remain neutral |
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Answer» Correct Answer - A Metal with `- ve E_(RP)^(@) ` or `+ve E_(OP)^(@)` possesses the tendency to get itself oxidised . |
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| 1195. |
Which of the following statement(s) is(are) correct ?A. The electrolysis of molten `NaOH` liberates `O_(2)` at the anodeB. The electrolysis of molten `NaOH` liberates `O_(2)` at the cathodeC. Molten `NaCl` conducts electricity due to the presence of `Na^(+)` and `Cl^(-)` ionsD. The electrolysis of molten `KCl` produces `Cl_(2)` at the cathode |
| Answer» Correct Answer - A::C | |
| 1196. |
Which of the following chages involves oxidation ?A. The conversion of ferrous sulphate to ferric sulphateB. The conversion of `H_(2)S` to `S`C. The conversion of `ZnSO_(4)` to `Zn`D. The conversion of `Zn` to `ZnSO_(4)` |
| Answer» Correct Answer - A::B::D | |
| 1197. |
Which of the following statements is (are) correct ?A. A metal in its highest oxidation state acts as an oxidantB. In the reaction, `F_(2) + (1)/(2)O_(2) rarr OF_(2)`, oxygen is an oxidantC. The oxidation number of `Ni` in `Ni(CO)_(4)` is zeroD. Copper metal can be oxidised by `Zn^(2+)` ions |
| Answer» Correct Answer - A::C | |
| 1198. |
The standard cell potential for the cell `Zn|Zn^(2+) (1 M)|| Cu^(2+) (1 M)| Cu` given `E_(Cu^(2+)//Cu)^(@) = 0.34 V` and `E_(Zn^(2+)//Zn)^(@) = - 0.76 V` isA. `-0.76 + 0.34 = - 0.42 V`B. `- 0.34 - (-0.76) = +0.42` VC. `0.34 - (-0.76) = + 1.10 V `D. `-0.76 - (+ 0.34) = -1.10 V` |
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Answer» Correct Answer - C `E_(cell)^(@) = E_(RP_(Cu))^(@) - E_(RP_(Zn))^(@)` `= 0.34 - (-0.76) = 1.10` V |
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| 1199. |
Which of the followinf statement(s) is(are) correct ?A. `Zn` is precipitated by the addition of `Cu` powder to a `ZnSO_(4)` solutionB. `AgNO_(3)` solution cen be stored in a copper containerC. When `Cl_(2)` is passed through `KBr(aq.)`, the solution becomes colouredD. The addition of a crystal of `I_(2)` turns a `KBr` solution violet |
| Answer» Correct Answer - C::D | |
| 1200. |
The equation representing the process by which standard reduction potential of zinc can be defined is:A. `Zn^(2+) (s) + 2e^(-) rarr Zn`B. `Zn (g) rarr Zn^(2+) (g) + 2e^(-)`C. `Zn^(2+) (g) + 2e^(-) rarr Zn`D. `Zn^(2+) (a.q) + 2e^(-) rarr Zn (s)` |
| Answer» Correct Answer - D | |