For the cell
Question:

For the cell

$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.1 \mathrm{M}) \| \mathrm{Ag}^{+}(\mathrm{aq})(0.01 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$

the cell potential $\mathrm{E}_{1}=0.3095 \mathrm{~V}$

For the cell

$\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(\mathrm{aq})(0.001 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$

the cell potential = $\times 10^{-2} \mathrm{~V}$. (Round off the Nearest Integer).

$\left[\right.$ Use $\left.: \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\right]$

Solution:

Cell reaction is :

$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$

Now, $\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {Cell }}^{o}-\frac{0.059}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}}$……..(1)

$\therefore \mathrm{E}_{1}=0.3095=\mathrm{E}_{\text {Cell }}^{\mathrm{o}}-\frac{0.059}{2} \cdot \log \frac{0.01}{(0.001)^{2}} \ldots$(2)

From (1) and (2) , $\mathrm{E}_{2}=0.28 \mathrm{~V}=28 \times 10^{-2} \mathrm{~V}$