Consider the Gravanic cell $Z n | Z n S O_{4} | | C u S O_{4} | C u$ the reaction at cathode is:

C u^{2 +} + 2 e^{-} \to C u

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Z n^{2 +} + 2 e^{-} \to Z n

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C u^{2 +} + Z n \to C u + Z n^{2 +}

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Z n^{2 +} + C u \to Z n + C u^{2 +}

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Solution

The correct option is A $C u^{2 +} + 2 e^{-} \to C u$
At the cathode reduction takes place and the half cell reaction is:

At anode: $C u^{2 +} + 2 e^{-} \to C u$

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Similar questions

For cell reaction, $Z n + C u^{2 +} \to Z n^{2 +} + C u,$ cell representation is

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

0.350

C u^{2 +} (a q) + 2 e^{-} \to C u

- 0.763

Z n^{2 +} (a q) + 2 e^{-} \to Z n

Z n + C u S O_{4} \to Z n S O_{4} + C u

E_{Z n / Z n^{2 +}}^{\circ} = 0.765 v o l t; E_{C u^{2 +} / C u}^{\circ} = 0.347 v o l t

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

C u^{2 +} (a q) + 2 e^{-} \to C u

Z n^{2 +} (a q) + 2 e^{-} \to Z n

The cell representation of the given reaction is:
Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)

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