Two students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)`. The `EMF` of one cell is `0.03` higher than the other. The concentration of `CuSO_(4)` in the cell with higher `EMF` value is `0.5M`. Find the concentration of `CuSO_(4)` in the other cell.
`(` Take `2.303 RT//F=0.06)`

Explanation

To solve the problem, we need to use the Nernst equation, which relates the EMF of a cell to the concentrations of the ions involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Nernst Equation**:
The Nernst equation for a cell can be expressed as:
\[ E = E^\circ - \frac{RT}{nF} \log \frac{[Ox]}{[Red]} \]
where:
- \( E \) is the cell potential (EMF),
- \( E^\circ \) is the standard cell potential,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( n \) is the number of moles of electrons exchanged,
- \( F \) is Faraday's constant,
- \([Ox]\) and \([Red]\) are the concentrations of the oxidized and reduced forms, respectively.

2. **Set Up the Equations for Both Cells**:
For Student 1 (higher EMF):
\[ E_1 = E^\circ - 0.06 \cdot \log \frac{C_1}{0.5} \]
For Student 2 (lower EMF):
\[ E_2 = E^\circ - 0.06 \cdot \log \frac{C_1}{C_2} \]
Here, \( C_2 \) is the concentration of \( CuSO_4 \) in the second cell, and \( C_1 \) is the concentration of \( ZnSO_4 \).

3. **Relate the Two EMFs**:
We know that \( E_1 - E_2 = 0.03 \):
\[ (E^\circ - 0.06 \cdot \log \frac{C_1}{0.5}) - (E^\circ - 0.06 \cdot \log \frac{C_1}{C_2}) = 0.03 \]
This simplifies to:
\[ -0.06 \cdot \log \frac{C_1}{0.5} + 0.06 \cdot \log \frac{C_1}{C_2} = 0.03 \]

4. **Factor Out Common Terms**:
\[ 0.06 \left( \log \frac{C_1}{C_2} - \log \frac{C_1}{0.5} \right) = 0.03 \]
This can be rewritten using the properties of logarithms:
\[ 0.06 \left( \log \frac{C_1 \cdot 0.5^{-1}}{C_2} \right) = 0.03 \]

5. **Simplify the Equation**:
Dividing both sides by \( 0.06 \):
\[ \log \frac{0.5}{C_2} = \frac{0.03}{0.06} = 0.5 \]

6. **Convert from Logarithmic to Exponential Form**:
\[ \frac{0.5}{C_2} = 10^{0.5} \]
Thus,
\[ C_2 = \frac{0.5}{10^{0.5}} = \frac{0.5}{\sqrt{10}} \approx \frac{0.5}{3.162} \approx 0.158 \]

7. **Final Calculation**:
Since \( 10^{0.5} \) is approximately \( 3.162 \):
\[ C_2 \approx \frac{0.5}{3.162} \approx 0.158 \text{ M} \]

### Final Answer:
The concentration of \( CuSO_4 \) in the other cell is approximately \( 0.158 \text{ M} \).