During the electrolysis of 0.1 M `CuSO_(4)` solution using coppepr electrodes, a depletion of `[Cu^(2+)]` occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the local concentration of `[Cu^(2+)]` near the anode and cathode are respectively 0.12 M and 0.08 M, calcualte the back emf developed. Temperature = 298 K.

Options

A. 22 mV
B. 5.2 mV
C. 29 mV
D. 59 mV

Explanation

To calculate the back EMF developed during the electrolysis of a 0.1 M CuSO₄ solution using copper electrodes, we will follow these steps:

### Step 1: Understand the Nernst Equation
The Nernst equation relates the cell potential to the concentrations of the reactants and products. It is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
where:
- \( E \) is the cell potential,
- \( E^\circ \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred,
- [products] and [reactants] are the concentrations of the products and reactants, respectively.

### Step 2: Determine the Reactions at the Electrodes
At the cathode, copper ions are reduced to copper metal:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \]
At the anode, copper metal is oxidized to copper ions:
\[ \text{Cu (s)} \rightarrow \text{Cu}^{2+} + 2e^- \]

### Step 3: Calculate the Potential at the Cathode
Using the Nernst equation for the cathode:
- \( E^\circ \) for copper is 0 V (standard reduction potential),
- \( n = 2 \) (2 electrons are transferred),
- The concentration of Cu²⁺ near the cathode is 0.08 M.

The Nernst equation for the cathode becomes:
\[ E_{\text{cathode}} = 0 - \frac{0.059}{2} \log \left( \frac{1}{0.08} \right) \]
Calculating this gives:
\[ E_{\text{cathode}} = -0.0295 \log(12.5) \]
\[ E_{\text{cathode}} = -0.0295 \times 1.096 \]
\[ E_{\text{cathode}} \approx -0.032 V \]

### Step 4: Calculate the Potential at the Anode
Using the Nernst equation for the anode:
- The concentration of Cu²⁺ near the anode is 0.12 M.

The Nernst equation for the anode becomes:
\[ E_{\text{anode}} = 0 - \frac{0.059}{2} \log \left( \frac{0.12}{1} \right) \]
Calculating this gives:
\[ E_{\text{anode}} = -0.0295 \log(0.12) \]
\[ E_{\text{anode}} = -0.0295 \times (-0.920) \]
\[ E_{\text{anode}} \approx 0.027 V \]

### Step 5: Calculate the Back EMF
The back EMF is the difference between the cathode and anode potentials:
\[ \text{Back EMF} = E_{\text{cathode}} - E_{\text{anode}} \]
Substituting the values:
\[ \text{Back EMF} = (-0.032) - (0.027) \]
\[ \text{Back EMF} = -0.005 V \]
Converting to millivolts:
\[ \text{Back EMF} = -5 \, \text{mV} \]

### Final Answer
The back EMF developed is approximately **5 mV**.

---