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The cell reaction for following galvanic cell is ${Pt.Cl}_{2}_{(g)} ∣ {Cl}^{-} (1 M) ∥ F^{-} (1 M) / F_{2 (g)}, Pt$

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${Cl}_{(aq)}^{-} + \frac{1}{2} F_{2 (g)} ⇌ \frac{1}{2} {Cl}_{2 (g)} + F_{(aq)}^{-}$

$\frac{1}{2} F_{2 (g)} + \frac{1}{2} {Cl}_{2 (g)} ⇌ F_{(g)}^{-} + {Cl}_{(g)}^{-}$

$F_{(aq)}^{-} + \frac{1}{2} {Cl}_{2 (g)} ⇌ {Cl}_{(aq)}^{-} + \frac{1}{2} F_{2 (g)}$

$F_{(aq)}^{-} + {Cl}_{(g)}^{-} ⇌ \frac{1}{2} F_{2 (g)} + \frac{1}{2} {Cl}_{2 (g)}$

answer is A.

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Detailed Solution

Non - metal electrodes

Higher SRP electrode under goes reduction

Lower SRP electrode under goes oxidation

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