`I_(2)(s)|I^(-)`(0.1M) half cell is connected to a `H^(+)` (aq)|`H_(2)`(1 bar)|Pt half celland e.m.f. is found to be 0.7714 V. If `E_(I_(2)|I^(-))^(@)=0.535 V`, find the pH of `H^(+)|H_(2)` half cell.

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the problem step by step, we will use the Nernst equation and the information provided in the question.

### Step 1: Identify the half-reactions
The half-cell reactions are:
- Cathode (reduction): \( I_2(s) + 2e^- \rightarrow 2I^-(aq) \)
- Anode (oxidation): \( H_2(g) \rightarrow 2H^+(aq) + 2e^- \)

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q \]
where:
- \( E_{cell} \) is the cell potential (0.7714 V)
- \( E^0_{cell} \) is the standard cell potential
- \( n \) is the number of moles of electrons transferred (2 in this case)
- \( Q \) is the reaction quotient

### Step 3: Calculate \( E^0_{cell} \)
The standard cell potential \( E^0_{cell} \) can be calculated as:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]
Given \( E^0_{I_2/I^-} = 0.535 V \) and \( E^0_{H_2/H^+} = 0 V \):
\[ E^0_{cell} = 0.535 V - 0 V = 0.535 V \]

### Step 4: Substitute values into the Nernst equation
Now substituting the known values into the Nernst equation:
\[ 0.7714 V = 0.535 V - \frac{0.0591}{2} \log Q \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 0.7714 V - 0.535 V = -\frac{0.0591}{2} \log Q \]
\[ 0.2364 V = -\frac{0.0591}{2} \log Q \]

### Step 6: Solve for \( \log Q \)
Multiply both sides by -2:
\[ -2 \times 0.2364 V = 0.0591 \log Q \]
\[ -0.4728 = 0.0591 \log Q \]
\[ \log Q = \frac{-0.4728}{0.0591} \approx -8.0 \]

### Step 7: Calculate \( Q \)
Now, converting from logarithmic form to exponential form:
\[ Q = 10^{-8.0} = 1.0 \times 10^{-8} \]

### Step 8: Write the expression for \( Q \)
The reaction quotient \( Q \) for the reaction is given by:
\[ Q = \frac{[H^+]^2 [I^-]^2}{[I_2][H_2]} \]
Since \( [I_2] \) is solid and its activity is 1, and \( [H_2] \) is 1 bar (or 1 atm), we have:
\[ Q = [H^+]^2 \times [I^-]^2 \]
Given \( [I^-] = 0.1 M \):
\[ Q = [H^+]^2 \times (0.1)^2 = [H^+]^2 \times 0.01 \]

### Step 9: Substitute \( Q \) back into the equation
From the previous step:
\[ 1.0 \times 10^{-8} = [H^+]^2 \times 0.01 \]
\[ [H^+]^2 = \frac{1.0 \times 10^{-8}}{0.01} = 1.0 \times 10^{-6} \]

### Step 10: Calculate \( [H^+] \)
Taking the square root:
\[ [H^+] = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3} M \]

### Step 11: Calculate pH
Finally, the pH is calculated as:
\[ pH = -\log[H^+] = -\log(1.0 \times 10^{-3}) = 3 \]

### Final Answer
The pH of the \( H^+/H_2 \) half-cell is **3**.

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