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$E_{{Sn}^{2 +} / Sn}^{\circ} = - 1.14 V, E_{{Pb}^{2 +} / Pb}^{\circ} = 0.126 V$
The cell reaction of the cell constructed from these two electrodes is

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Pb+Sn→Pb²⁺+Sn²⁺

Pb²⁺+Sn²⁺→Pb+Sn

Sn²⁺+Pb→Sn+Pb²⁺

Sn+Pb²⁺→Sn²⁺+Pb

answer is D.

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Detailed Solution

$E_{{Sn}^{2 +} / Sn}^{\circ} = - 1.14 V \Rightarrow$ anode

$E_{{Pb}^{2 +} / Pb}^{\circ} = 0.126 V \Rightarrow$ cathode

$\Rightarrow Sn \to {Sn}^{2 +} + 2 e^{-} {Pb}^{2 +} + 2 e^{-} \to Pb__________________{Pb}^{2 +} + Sn \to Pb + {Sn}^{2 +}$

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