$Z n (s) | Z n {C l}_{2} (a q) | ∣ ∣ {C l}^{-} (a q) ∣ ∣ | {C l}_{2} (g) | C (s)$
The reaction at anode in the electrochemical cell is:

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

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

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{C l}_{2} + 2 e \to 2 {C l}^{-}

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2 {C l}^{-} \to {C l}_{2} + 2 e

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Z n + {C l}_{2} \to Z n {C l}_{2}

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Solution

The correct option is A $Z n \to {Z n}^{2 +} + 2 e$
The reaction at anode in the electrochemical cell is
$Z n \to {Z n}^{2 +} + 2 e$
It is an oxidation reaction
The reaction at cathode in the electrochemical cell is
${C l}_{2} + 2 e \to 2 {C l}^{-}$
It is a reduction reaction.
In the cell notation, the left hand electrode is anode where oxidation half reaction occurs.
The right hand electrode is cathode where reduction half reaction occurs.

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Similar questions

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

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

Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

Z n (s) ⟶ {Z n}^{2 +} (a q) + 2 e^{-}

{P b}^{2 +} (a q) + 2 e^{-} ⟶ P b (s)

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 \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

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

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