Calcualte the cell EMP in mV for
`Pt|H_(2)(1 atm) | HCl(0.01 M)|| AgCl(s)| Ag(s)` at 298 K
If `DeltaG_(f)^(@)` values are at `25^(@)C`,
`-109.56 (kJ)/(mol) for AgCl(s)` and
`-130.79 (kJ)/(mol)` for `(H^(+) + Cl^(-)(aq)`, Take 1F = 96500 C

Options

A. 456 mV
B. 654 mV
C. 646 mV
D. None of these

Explanation

To calculate the cell EMF for the given electrochemical cell setup, we will follow these steps:

### Step 1: Write the balanced half-reactions
For the cell `Pt|H2(1 atm) | HCl(0.01 M)|| AgCl(s)| Ag(s)`, we identify the half-reactions:

- **Anode (oxidation)**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]

- **Cathode (reduction)**:
\[ AgCl(s) + e^- \rightarrow Ag(s) + Cl^-(aq) \]

### Step 2: Combine the half-reactions
To combine the half-reactions, we need to balance the electrons. The oxidation half-reaction produces 2 electrons, while the reduction half-reaction consumes 1 electron. Thus, we multiply the reduction half-reaction by 2:

\[ 2AgCl(s) + 2e^- \rightarrow 2Ag(s) + 2Cl^-(aq) \]

Now, the overall balanced equation is:
\[ H_2(g) + 2AgCl(s) \rightarrow 2H^+(aq) + 2Ag(s) + 2Cl^-(aq) \]

### Step 3: Calculate ΔG° for the reaction
Using the given ΔG° values:
- For \( H^+ + Cl^- \): \( \Delta G_f^\circ = -130.79 \, \text{kJ/mol} \)
- For \( AgCl \): \( \Delta G_f^\circ = -109.56 \, \text{kJ/mol} \)

The ΔG° for the reaction can be calculated as follows:
\[ \Delta G^\circ_{\text{cell}} = \sum \Delta G^\circ_{\text{products}} - \sum \Delta G^\circ_{\text{reactants}} \]

Substituting the values:
\[ \Delta G^\circ_{\text{cell}} = [2 \times (-130.79) + 0] - [0 + 2 \times (-109.56)] \]
\[ = -261.58 + 219.12 = -42.46 \, \text{kJ/mol} \]

### Step 4: Convert ΔG° to Joules
Convert ΔG° from kJ to J:
\[ \Delta G^\circ_{\text{cell}} = -42460 \, \text{J/mol} \]

### Step 5: Calculate the cell potential (E°cell)
Using the relationship:
\[ \Delta G^\circ_{\text{cell}} = -nFE^\circ_{\text{cell}} \]
Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, \text{C/mol} \)

Rearranging for \( E^\circ_{\text{cell}} \):
\[ E^\circ_{\text{cell}} = -\frac{\Delta G^\circ_{\text{cell}}}{nF} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = -\frac{-42460}{2 \times 96500} \]
\[ = \frac{42460}{193000} \approx 0.220 \, \text{V} \]

### Step 6: Apply the Nernst equation
Now we apply the Nernst equation to find the cell potential under non-standard conditions:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \]
Where \( Q \) is the reaction quotient.

For the reaction:
\[ Q = \frac{[H^+]^2}{P_{H_2} \cdot [AgCl]^2} \]
Given:
- \( [H^+] = 0.01 \, \text{M} \)
- \( P_{H_2} = 1 \, \text{atm} \)
- \( [AgCl] = 1 \, \text{(solid, activity = 1)} \)

Thus:
\[ Q = \frac{(0.01)^2}{1 \cdot 1} = 0.0001 \]

Substituting into the Nernst equation:
\[ E_{\text{cell}} = 0.220 - \frac{0.059}{2} \log(0.0001) \]
Calculating \( \log(0.0001) = -4 \):
\[ E_{\text{cell}} = 0.220 - \frac{0.059}{2} \times (-4) \]
\[ = 0.220 + 0.118 = 0.338 \, \text{V} \]

### Step 7: Convert to millivolts
Finally, converting to millivolts:
\[ E_{\text{cell}} = 0.338 \times 1000 = 338 \, \text{mV} \]

### Final Answer
The cell EMF is approximately **338 mV**.

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