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.

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

1. A

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

2. B

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

3. C

more of Fe and Pb

4. D

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

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### Solution:

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

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