The EMF of the cell `M|M^(n+)(0.02M) ||H^(+)(1M)|H_(2)(g) (1atm)Pt` at `25^(@)C` is `0.81V`. Calculate the valency of the metal if the standard oxidation potential of the metal is `0.76V`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions
From the cell representation \( M | M^{n+} (0.02M) || H^+ (1M) | H_2 (g) (1atm) Pt \), we can identify the half-reactions:
- **Anode (oxidation)**: \( M \rightarrow M^{n+} + ne^- \)
- **Cathode (reduction)**: \( 2H^+ + 2e^- \rightarrow H_2 \)

### Step 2: Write the net cell reaction
To balance the number of electrons transferred, we multiply the cathode reaction by \( n \) and the anode reaction by \( 2 \):
- **Net Cell Reaction**:
\[ 2M + 2nH^+ \rightarrow 2M^{n+} + nH_2 \]

### Step 3: Apply the Nernst equation
The Nernst equation at room temperature (25°C) is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
Where:
- \( E_{cell} = 0.81V \)
- \( E^0_{cell} = E^0_{anode} + E^0_{cathode} = 0.76V + 0V = 0.76V \)
- The concentration of \( M^{n+} \) is \( 0.02M \), and the pressure of \( H_2 \) is \( 1 atm \).
- The concentration of \( H^+ \) is \( 1M \).

### Step 4: Substitute values into the Nernst equation
Substituting the known values into the Nernst equation:
\[ 0.81 = 0.76 - \frac{0.0591}{n} \log \left( \frac{(0.02)^n \cdot (1)^n}{(1)^{2n}} \right) \]
This simplifies to:
\[ 0.81 = 0.76 - \frac{0.0591}{n} \log \left( 0.02^n \right) \]

### Step 5: Solve for \( n \)
Rearranging gives:
\[ 0.81 - 0.76 = - \frac{0.0591}{n} \log(0.02^n) \]
\[ 0.05 = - \frac{0.0591}{n} \cdot n \log(0.02) \]
\[ 0.05 = -0.0591 \log(0.02) \]
Calculating \( \log(0.02) \):
\[ \log(0.02) \approx -1.699 \]
Thus:
\[ 0.05 = 0.0591 \cdot 1.699 \]
Calculating gives:
\[ 0.05 = 0.1004 \]
Now we can find \( n \):
\[ n = \frac{0.0591 \cdot 1.699}{0.05} \]
Calculating gives:
\[ n \approx 2 \]

### Conclusion
The valency of the metal \( M \) is \( n = 2 \).

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