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JEE Main 20204 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2020 · 4 September · Morning Shift · Chemistry · Electrochemistry

E _ Cu ^ 2 + | Cu ^ 0 = + 0 .34 V E _ Zn ^ 2 + | Zn ^ 0 = - 0 .76 V...

E _ Cu ^ 2 + | Cu ^ 0 = + 0 .34 V E _ Zn ^ 2 + | Zn ^ 0 = - 0 .76 V Identify the incorrect statement from the options
ECu2+∣Cu0=+0.34V{E}_{{Cu}^{2 +} | Cu}^{0} = + 0 .34 V
EZn2+∣Zn0=−0.76V{E}_{{Zn}^{2 +} | Zn}^{0} = - 0 .76 V
Identify the incorrect statement from the options below for the above cell:

Options

  1. AIf Eext>1.1V,e–{E}_{ext} > 1 .1 V, {e}^{–} flow from CuCu to ZnZn
  2. BIf Eext>1.1V,Zn{E}_{ext} > 1 .1 V, Zn dissolves at ZnZn electrode and CuCu deposits at CuCu electrode
  3. CIf Eext<1.1V,Zn{E}_{ext} < 1 .1 V, Zn dissolves at anode and CuCu deposits at cathode
  4. DIf Eext=1.1V,{E}_{ext} = 1 .1 V, no flow of e–{e}^{–} or current occurs

Correct answer

B. If Eext>1.1V,Zn{E}_{ext} > 1 .1 V, Zn dissolves at ZnZn electrode and CuCu deposits at CuCu electrode

Step-by-step solution

{E}_{cell}^{∘} = {E}_{{Cu}^{2 +}\left|Cu\right\}^{∘} - {E}_{{Zn}^{2 +}\left|Zn\right\}^{∘} = 1 .1 V

So If E=1.1VE = 1 .1 V no electron will flow

At E>1.1VE > 1 . 1V cell act as electrolytic cell and electron will flow from CuCu to ZnZn.

At E<1.1VE < 1 . 1V cell act as electrochemical so ZnZn dissolve and CuCu deposit.

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