Construct at cell consisting of `Ni_((ag))^(2+)| Ni_((s))` half cell and `H^(+) ||H_(2(g)) | Pt` half cell
(a) Write the cell reaction
( b) Calculate emf of the cell if cell `[Ni^(2+)] = 0.1 `
`[ H^(+)] =0.05 M` and `E_(Ni)^(0) =- 0.257 V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the half-cell reactions
We have two half-cells:
1. The nickel half-cell:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}_{(s)} \quad (E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.257 \, \text{V}) \]
2. The hydrogen half-cell:
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2_{(g)} \quad (E^\circ_{\text{H}^+/H_2} = 0 \, \text{V}) \]

### Step 2: Write the overall cell reaction
In the overall cell reaction, nickel is oxidized and hydrogen ions are reduced:
\[ \text{Ni}_{(s)} + 2\text{H}^+ \rightarrow \text{Ni}^{2+} + \text{H}_2_{(g)} \]

### Step 3: Calculate the standard cell potential \(E^\circ_{\text{cell}}\)
The standard cell potential is calculated by combining the reduction potential of the cathode and the oxidation potential of the anode:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0 \, \text{V} - (-0.257 \, \text{V}) = 0.257 \, \text{V} \]

### Step 4: Use the Nernst equation to calculate the cell potential
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- \(n = 2\) (the number of electrons transferred)
- \([\text{products}] = [\text{Ni}^{2+}] = 0.1 \, \text{M}\)
- \([\text{reactants}] = [\text{H}^+] = 0.05 \, \text{M}\) and the pressure of \(\text{H}_2\) is 1 atm.

Substituting into the Nernst equation:
\[ E_{\text{cell}} = 0.257 - \frac{0.0591}{2} \log \left( \frac{0.1}{0.05} \right) \]

### Step 5: Calculate the logarithm
\[ \log \left( \frac{0.1}{0.05} \right) = \log(2) \approx 0.301 \]

### Step 6: Substitute back into the Nernst equation
\[ E_{\text{cell}} = 0.257 - \frac{0.0591}{2} \times 0.301 \]
\[ E_{\text{cell}} = 0.257 - 0.0296 \times 0.301 \]
\[ E_{\text{cell}} = 0.257 - 0.0089 \approx 0.2481 \, \text{V} \]

### Final Answer
The emf of the cell is approximately:
\[ E_{\text{cell}} \approx 0.2481 \, \text{V} \]

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