For the cell Cu(s)|Cu$$^{2+}$$(aq)(0.1M)||Ag$$^+$$(aq)(0.01M)|Ag(s) the cell potential E$$_1$$ = 0.3095 V. For the cell Cu(s)|Cu$$^{2+}$$(aq)(0.01M)||Ag$$^+$$(aq)(0.001M)|Ag(s) the cell potential = $$x \times 10^{-2}$$ V. Find value of x.
(Round off to the Nearest Integer).
[Use: $$\frac{2.303 RT}{F} = 0.059$$ J]

The cell is written as Cu(s)|Cu$$^{2+}$$(aq)||Ag$$^+$$(aq)|Ag(s). The electrode on the left (copper) undergoes oxidation while the silver electrode on the right undergoes reduction, so the overall cell reaction is

$$\text{Cu}(s)+2\,\text{Ag}^+(aq)\;\longrightarrow\;\text{Cu}^{2+}(aq)+2\,\text{Ag}(s)$$

For any cell reaction the Nernst equation at 298 K is first stated as

$$E = E^\circ - \frac{0.059}{n}\,\log Q,$$

where $$E^\circ$$ is the standard cell potential, $$n$$ is the number of electrons transferred and $$Q$$ is the reaction quotient. Here $$n = 2$$ (two electrons are exchanged) and

$$Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^{\,2}}.$$

We are given the potential of the first cell,

$$E_1 = 0.3095\ \text{V}$$

when $$[\text{Cu}^{2+}] = 0.1\ \text{M}$$ and $$[\text{Ag}^+] = 0.01\ \text{M}$$. We therefore compute

$$Q_1 = \frac{0.1}{(0.01)^{2}} = \frac{0.1}{1.0\times10^{-4}} = 1.0\times10^{3},$$

so that

$$\log Q_1 = \log(1.0\times10^{3}) = 3.$$

Substituting these values in the Nernst equation,

$$0.3095 = E^\circ - \frac{0.059}{2}\times 3,$$

$$E^\circ = 0.3095 + \frac{0.059}{2}\times 3,$$

$$E^\circ = 0.3095 + 0.0885 = 0.398\ \text{V}.$$

Now we need the potential of the second cell, for which

$$[\text{Cu}^{2+}] = 0.01\ \text{M},\qquad [\text{Ag}^+] = 0.001\ \text{M}.$$

Thus

$$Q_2 = \frac{0.01}{(0.001)^{2}} = \frac{0.01}{1.0\times10^{-6}} = 1.0\times10^{4},$$

and

$$\log Q_2 = \log(1.0\times10^{4}) = 4.$$

Applying the Nernst equation again,

$$E_2 = E^\circ - \frac{0.059}{2}\times 4,$$

$$E_2 = 0.398 - 0.0295 \times 4,$$

$$E_2 = 0.398 - 0.118 = 0.280\ \text{V}.$$

This potential is written in the form $$E_2 = x\times10^{-2}\ \text{V}$$, hence

$$0.280\ \text{V} = x\times10^{-2}\ \text{V}\quad\Longrightarrow\quad x = 28.$$

Hence, the correct answer is Option 28.