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Calculate $\mathrm{E}_{\text {cell }}^{\circ}$ for the following.
$\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(1 \mathrm{M})}^{++}\right|\left|\mathrm{Pb}_{(1 \mathrm{M})}^{++}\right| \mathrm{Pb}_{(\mathrm{s})}$ if $\mathrm{E}_{\mathrm{Zn}^{\circ}}^{\circ}=-0.763 \mathrm{~V}$ and
$\mathrm{E}_{\mathrm{Pb}}^{\mathrm{o}}=-0.126 \mathrm{~V}$
Options:
$\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(1 \mathrm{M})}^{++}\right|\left|\mathrm{Pb}_{(1 \mathrm{M})}^{++}\right| \mathrm{Pb}_{(\mathrm{s})}$ if $\mathrm{E}_{\mathrm{Zn}^{\circ}}^{\circ}=-0.763 \mathrm{~V}$ and
$\mathrm{E}_{\mathrm{Pb}}^{\mathrm{o}}=-0.126 \mathrm{~V}$
Solution:
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Verified Answer
The correct answer is:
$0.637 \mathrm{~V}$
For the given cell, $\mathrm{Zn}$ acts as anode and $\mathrm{Pb}$ acts as cathode.
$\mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0$
$=\mathrm{E}_{\mathrm{Pb}}^0-\mathrm{E}_{\mathrm{Zn}}^0=-0.126 \mathrm{~V}-(-0.763) \mathrm{V}$
$=0.637 \mathrm{~V}$
$\mathrm{E}_{\text {cell }}^0=\mathrm{E}_{\text {cathode }}^0-\mathrm{E}_{\text {anode }}^0$
$=\mathrm{E}_{\mathrm{Pb}}^0-\mathrm{E}_{\mathrm{Zn}}^0=-0.126 \mathrm{~V}-(-0.763) \mathrm{V}$
$=0.637 \mathrm{~V}$
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