E Cu 2 + | Cu 0 = + 0 .34   V E Zn 2 + | Zn 0 = − 0 .76   V Identify the incorrect statement…


ECu2+|Cu0=+0.34 V
EZn2+|Zn0=0.76 V
Identify the incorrect statement from the options below for the above cell:
  1. If Eext>1.1 V, e flow from Cu to Zn
  2. If Eext>1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
  3. If Eext<1.1 V, Zn dissolves at anode and Cu deposits at cathode
  4. If Eext=1.1 V, no flow of e or current occurs

Solution

Ecell=ECu2+CuEZn2+Zn=1.1V

So If E=1.1V no electron will flow

At E>1.1 V cell act as electrolytic cell and electron will flow from Cu to Zn.

At E<1.1 V cell act as electrochemical so Zn dissolve and Cu deposit.

Asked in: JEE Main 2020 (04 Sep Shift 1)

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