For the redox reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) ightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$ taking place in a cell, $\mathrm{E}_{ ext {cell }}^0$ is $1.10$ volt. $\mathrm{E}^{\circ}$ for the cell will be $\left(2.303…

Question

For the redox reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) ightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$ taking place in a cell, $\mathrm{E}_{	ext {cell }}^0$ is $1.10$ volt. $\mathrm{E}^{\circ}$ for the cell will be $\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.0591ight)$

Aicharya · Accepted Answer

$\mathrm{E}_{	ext {cell }}=\mathrm{E}_{	ext {cell }}^0+\frac{0.059}{\mathrm{n}} \log \left[\frac{\mathrm{Cu}^{+2}}{\mathrm{Zn}^{+2}}ight]$ $=1.10+\frac{0.059}{2} \log [0.1]=1.10-0.0295=1.07 \mathrm{~V}$

For the redox reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightarrow…

Solution