For the galvanic cell,

Question:

For the galvanic cell,

$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$

$\mathrm{E}_{\mathrm{cell}}=\longrightarrow \times 10^{-2} \mathrm{~V} .$ (Nearest integer)

$\left[\right.$ Use : $\mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{0}=-0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{0}=+0.76 \mathrm{~V},$,

$\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right]$

Solution:

Galvanic cell:

$\mathrm{Zn}_{(\mathrm{s})}+\underset{0.02 \mathrm{M}}{\mathrm{Cu}_{(\mathrm{aq} .)}^{+2}} \rightarrow \underset{0.04 \mathrm{M}}{\mathrm{Zn}^{+2}}+\mathrm{Cu}(\mathrm{s})$

Nernst equation $=\mathrm{F}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\frac{0.059}{2} \log \frac{\left[2 \mathrm{n}^{+2}\right]}{\left[\mathrm{Cu}^{+2}\right]}$

$\Rightarrow \mathrm{E}_{\text {cell }}\left[\mathrm{E}_{\text {cell }}^{\mathrm{o}}-\mathrm{E}_{\mathrm{Zn}^{+2} / \mathrm{Zn}}^{\mathrm{o}}\right]-\frac{0.059}{2} \log \frac{0.04}{0.02}$

$\Rightarrow \mathrm{E}_{\text {cell }}[0.34-(-0.76)]-\frac{0.059}{2} \log ^{2}$

$\Rightarrow \mathrm{E}_{\text {cell }} 1-1-\frac{0.059}{2} \times 0.3010$

$=1.0911=109.11 \times 10^{-2}$

$=109$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now