Adding powdered Pb and Fe to a solution containing 1 M in each of Pb2+ and Fe2+ ions would result into the formation of Given: E∘Pb2+∣Pb=−0.126Vand E∘Fe2+∣Fe=−0.44V.

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Adding powdered Pb and Fe to a solution containing 1 M in each of ${Pb}^{2 +}$ and ${Fe}^{2 +}$ ions would result into the formation of Given: $E^{\circ} ({Pb}^{2 +} ∣ Pb) = - 0.126 V$ and $E^{\circ} ({Fe}^{2 +} ∣ Fe) = - 0.44 V)$ .

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more of Fe and Pb

more of ${Fe}^{2 +}$ and ${Pb}^{2 +}$ ions

more of Pb and ${Fe}^{2 +}$ ions

more of Fe and ${Pb}^{2 +}$ ions

answer is A.

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Detailed Solution

${Pb}^{2 +}$ will reduce to Pb and Fe will be oxidize to ${Fe}^{2 +}$ so as to have $E^{\circ}$ positive for the reaction $Pb + {Fe}^{2 +}$ . Hence, more of Pb and ${Fe}^{2 +}$ ions are formed.