The potential of the cell `Cu, Cu^(2+) (0.1M)"//"HCl (xM), Cl_(2), Pt`is 1.07 V. If the standard potential of copper and chlorine electrodes are 0.34 V and 1.36 V, calculate the concentration of HCI.

Answer

To solve the problem, we need to calculate the concentration of HCl (x M) in the given electrochemical cell. We will follow these steps:

### Step 1: Identify the half-reactions and standard potentials
The half-reactions for the cell are:
- At the cathode (reduction): \( \text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^- \) (Standard potential \( E^\circ_{\text{cathode}} = 1.36 \, \text{V} \))
- At the anode (oxidation): \( \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \) (Standard potential \( E^\circ_{\text{anode}} = 0.34 \, \text{V} \))

### Step 2: Calculate the standard cell potential
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 1.36 \, \text{V} - 0.34 \, \text{V} = 1.02 \, \text{V} \]

### Step 3: Use the Nernst equation
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
For our cell, the equation becomes:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \frac{[\text{Cl}^-]^2}{[\text{Cu}^{2+}]} \]
Given that \( E_{\text{cell}} = 1.07 \, \text{V} \) and \( [\text{Cu}^{2+}] = 0.1 \, \text{M} \):
\[ 1.07 = 1.02 - \frac{0.059}{2} \log \frac{x^2}{0.1} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 1.07 - 1.02 = -\frac{0.059}{2} \log \frac{x^2}{0.1} \]
\[ 0.05 = -\frac{0.059}{2} \log \frac{x^2}{0.1} \]

### Step 5: Solve for the logarithm
Multiplying both sides by -2:
\[ -0.1 = 0.059 \log \frac{x^2}{0.1} \]
Dividing by 0.059:
\[ \log \frac{x^2}{0.1} = -\frac{0.1}{0.059} \approx -1.6949 \]

### Step 6: Exponentiate to solve for x
Now we can solve for \( x^2 \):
\[ \frac{x^2}{0.1} = 10^{-1.6949} \]
Calculating \( 10^{-1.6949} \):
\[ x^2 = 0.1 \times 0.0203 \approx 0.00203 \]
Taking the square root:
\[ x \approx \sqrt{0.00203} \approx 0.045 \]

### Step 7: Final concentration of HCl
Thus, the concentration of HCl is approximately:
\[ x \approx 0.045 \, \text{M} \]