For the cell : `Zn//Zn^(2+)` (x M) | | `Ag^(+)`/Ag (y M)
(a) Write Nernst Equation to show that how E(cell) vary with concentration of `Zn^(2+)` and `Ag^(+)` ions.
(b) Find E(cell) for `[Zn^(2+)]` = 0.01M and `[Ag^+]` = 0.05 M

Answer

To solve the problem, we will break it down into two parts as requested:

### Part (a): Write the Nernst Equation

1. **Identify the half-reactions**:
- For the reduction of silver ions:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^\circ = 0.80 \, \text{V}) \]
- For the oxidation of zinc:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad (E^\circ = -0.76 \, \text{V}) \]

2. **Determine the overall cell reaction**:
\[ \text{Zn} + 2\text{Ag}^+ \rightarrow \text{Zn}^{2+} + 2\text{Ag} \]

3. **Calculate the standard cell potential (\(E^\circ_{\text{cell}}\))**:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 \, \text{V} - (-0.76 \, \text{V}) = 1.56 \, \text{V} \]

4. **Write the Nernst equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \(R\) = universal gas constant = 8.314 J/(mol·K)
- \(T\) = temperature in Kelvin (assume 298 K)
- \(n\) = number of moles of electrons transferred (2 for this reaction)
- \(F\) = Faraday's constant = 96500 C/mol
- \(Q\) = reaction quotient = \(\frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2}\)

5. **Substituting values**:
The Nernst equation becomes:
\[ E_{\text{cell}} = 1.56 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} \right) \]

### Part (b): Calculate \(E_{\text{cell}}\) for \([\text{Zn}^{2+}] = 0.01 \, \text{M}\) and \([\text{Ag}^+] = 0.05 \, \text{M}\)

1. **Substitute the concentrations into the Nernst equation**:
\[ E_{\text{cell}} = 1.56 \, \text{V} - \frac{0.0591}{2} \log \left( \frac{0.01}{(0.05)^2} \right) \]

2. **Calculate the reaction quotient \(Q\)**:
\[ Q = \frac{0.01}{(0.05)^2} = \frac{0.01}{0.0025} = 4 \]

3. **Calculate the logarithm**:
\[ \log(4) \approx 0.602 \]

4. **Substitute back into the equation**:
\[ E_{\text{cell}} = 1.56 \, \text{V} - \frac{0.0591}{2} \times 0.602 \]
\[ E_{\text{cell}} = 1.56 \, \text{V} - 0.0178 \approx 1.5806 \, \text{V} \]

### Final Answer
\[ E_{\text{cell}} \approx 1.5806 \, \text{V} \]