Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Electrochemistry question

2019 · 9 Jan · Shift 2 · Q11
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Electrochemistry
  5. /2019 · 9 Jan · Shift 2 · Q11

Electrochemistry question

2019 · 9 Jan · Shift 2 · Q11

JEE MainChemistryElectrochemistryMCQ+4 / −1
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction ZnZnZn(s) + Cu2+Cu^{2+}Cu2+ (aq) ⇌\rightleftharpoons⇌ Zn2+Zn^{2+}Zn2+(aq) + CuCuCu(s) at 300 K is approximately, (R = 8 JK −-− 1mol −-− 1, F = 96000 C mol −-− 1)
  1. A
    e −-− 80
  2. B
    e −-− 160
  3. C
    e320
  4. D
    e160
View written solutionFree

Correct answer: D

  1. Use the relation between standard cell potential and equilibrium constant

For a cell reaction at equilibrium,

ΔG∘=−nFE∘\Delta G^\circ = -n F E^\circΔG∘=−nFE∘

and also,

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln KΔG∘=−RTlnK

Equating both,

−nFE∘=−RTln⁡K-nFE^\circ = -RT \ln K−nFE∘=−RTlnK

So,

ln⁡K=nFE∘RT\ln K = \frac{nFE^\circ}{RT}lnK=RTnFE∘​

  1. Identify the number of electrons transferred

Reaction:

Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s)Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s)

Half-reactions are:

Zn→Zn2++2e−Zn \rightarrow Zn^{2+} + 2e^-Zn→Zn2++2e− Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow CuCu2++2e−→Cu

Thus,

n=2n = 2n=2

  1. Substitute the values

Given:

  • E∘=2 VE^\circ = 2\,VE∘=2V
  • T=300 KT = 300\,KT=300K
  • R=8 J K−1mol−1R = 8\,J\,K^{-1}mol^{-1}R=8JK−1mol−1
  • F=96000 C mol−1F = 96000\,C\,mol^{-1}F=96000Cmol−1

Now,

ln⁡K=(2)(96000)(2)(8)(300)\ln K = \frac{(2)(96000)(2)}{(8)(300)}lnK=(8)(300)(2)(96000)(2)​

ln⁡K=3840002400=160\ln K = \frac{384000}{2400} = 160lnK=2400384000​=160

Therefore,

K=e160K = e^{160}K=e160

  1. Match with the given options

The correct option is:

e160\boxed{e^{160}}e160​

So, Option D is correct.

PreviousNext

More from Electrochemistry

  • Consider the statements S1 and S2 S1 : Conductivity always increases with decrease in the concentration of electrolyte. S2 : Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among…2019 · MCQ
  • Which one of the following graphs between molar conductivity (Λm​) versus C​ is correct ?2019 · MCQ
  • Consider the following reduction processes : Zn2+ + 2e– → Zn(s) ; Eo = – 0.76 V Ca2+ + 2e– → Ca(s); Eo = –2.87 V Mg2+ + 2e– → Mg(s) ; Eo = – 2.36 V Ni2​ + 2e– → Ni(s) ; Eo = – 0.25 The reducing power of the…2019 · MCQ
  • In the cell Pt ∣(s)∣ H2​(g, 1 bar) ∣HCl(aq)∣ AgCl ∣(s)∣ Ag(s)|Pt(s) the cell potential is 0.92 V when a 10–6 molal HCl solution is used.…2019 · MCQ
  • For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below : If Ezn2+/zn0​=− 0.76 V, which cathode will give maximum value of Eocell per electron transferred? Includes table2019 · MCQ
  • Given the equilibrium constant: KC of the reaction : Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) is 10 × 1015, calculate the E cell0​ of this reaciton at 298 K [2.303 FRT​ at 298 K = 0.059V]2019 · MCQ
  • Given CO3+​ + e– → CO2+ ; Eo = + 1.81 V Pb4+ + 2e– → Pb2+ ; Eo = + 1.67 V Ce4+ + e– → Ce3+ ; Eo = + 1.61 V Bi3+ + 3e– → Bi ; Eo = + 0.20 V Oxidizing power of the species will increase in…2019 · MCQ
  • The standard electrode potential Eo and its temperature coefficient (dTdEo​) for a cell are 2V and − 5 × 10 − 4 VK − 1 at 300 K respectively. The cell reaction is Zn(s) + Cu2+…2019 · MCQ