$$\begin{aligned} & \mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \end{aligned}$$ Identify the incorrect statement from the options below for the above cell:

(C) If $\mathrm{E}_{\text {ext }}>1.1 \mathrm{~V}$, $\mathrm{Zn}$ dissolves at Zn electrode and Cu deposits at Cu electrode

(A) If $\mathrm{E}_{\text {ext }}<1.1 \mathrm{~V}$, $\mathrm{Zn}$ dissolves at anode and Cu deposits at cathode

(B) If $\mathrm{E}_{\text {ext }}=1.1 \mathrm{~V}$, no flow of $\mathrm{e}^{-}$ or current occurs

(D) If $\mathrm{E}_{\mathrm{ext}}>1.1 \mathrm{~V}$, $\mathrm{e}^{-}$ flows from Cu to Zn

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Redox Reactions and Electrochemistry

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E_{Cu^{2 +} ∣ Cu}^{\circ} = + 0.34 V E_{Zn^{2 +} ∣ Zn}^{\circ} = - 0.76 V

Identify the incorrect statement from the options below for the above cell:

(A)If $E_{ext} < 1.1 V$ , $Zn$ dissolves at anode and Cu deposits at cathode
(B)If $E_{ext} = 1.1 V$ , no flow of $e^{-}$ or current occurs
(C)If $E_{ext} > 1.1 V$ , $Zn$ dissolves at Zn electrode and Cu deposits at Cu electrode✓ Correct
(D)If $E_{ext} > 1.1 V$ , $e^{-}$ flows from Cu to Zn

Answer: (C)

Solution

For this cell

E_{cell}^{\circ} = 0.34 - (- 0.76) = 1.1

V. When an opposing external voltage larger than

1.1

V is applied, it overpowers the cell and drives the reaction backwards, turning it into an electrolytic cell: copper is now oxidised (it becomes the anode) and zinc is deposited (it becomes the cathode). The statement that, for

E_{ext} > 1.1

V, zinc still dissolves at the zinc electrode while copper deposits at the copper electrode describes the ordinary spontaneous direction, not the reversed one, so it is the incorrect statement.

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Redox Reactions and Electrochemistry

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