Given E∘Cr3+∣Cr=−0.72V and  E∘Fe2+∣Fe=−0.42V The potential of the cell CrCr3+(0.1M)∥Fe2+(0.01M)Fe is

# Given ${E}^{\circ }\left({\mathrm{Cr}}^{3+}\mid \mathrm{Cr}\right)=-0.72\mathrm{V}$ and  The potential of the cell $\mathrm{Cr}\left|{\mathrm{Cr}}^{3+}\left(0.1\mathrm{M}\right)\parallel {\mathrm{Fe}}^{2+}\left(0.01\mathrm{M}\right)\right|\mathrm{Fe}$ is

1. A

2. B

3. C

4. D

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### Solution:

The cell potential is given by  where

Right half-cell  ${\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Fe}$

Left half-cell  ${\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Cr}$

Hence, ${E}_{\mathrm{cell}}=-0.48\mathrm{V}-\left(-0.74\mathrm{V}\right)=0.26\mathrm{V}$

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