`underset(1 L solution)(Ag|Ag^(o+)(1M))||underset(1 L solution)(Ag^(o+)||Ag)`
`0.5F` electricity in the `LHS(` anode `)` and `1F` of electricity in the `RHS(` cathode`)` is first passed making them independent electrolytic cells at `298K`. `EMF` of the cell after electrolysis will be
(a)Increased
(b)Decreased
(c)No change
(d)Time is also required

Options

A. Increased
B. Decreased
C. No change
D. Time is also required

Explanation

To solve the problem, we need to analyze the electrochemical cell described and determine how the EMF (Electromotive Force) changes after passing the specified amounts of electricity through the anode and cathode.

### Step-by-Step Solution:

1. **Understanding the Cell Configuration**:
- We have two half-cells:
- Left-hand side (LHS): Ag | Ag⁺ (1 M)
- Right-hand side (RHS): Ag⁺ | Ag (1 M)
- The LHS is the anode and the RHS is the cathode.

2. **Initial EMF Calculation**:
- The standard EMF (E°) of the cell can be calculated using the Nernst equation:
\[ E = E^\circ + \frac{0.0591}{n} \log \left( \frac{[Ag^+]_{RHS}}{[Ag^+]_{LHS}} \right) \]
- Here, \( n = 1 \) (1 electron is involved in the silver ion reduction/oxidation).

3. **Initial Concentrations**:
- Before passing any current, the concentrations of Ag⁺ in both half-cells are 1 M.
- Therefore, the initial EMF can be calculated as:
\[ E = 0 + \frac{0.0591}{1} \log \left( \frac{1}{1} \right) = 0 \text{ volts} \]

4. **Passing Electricity**:
- 0.5 F of electricity is passed through the LHS (anode), which will deposit 0.5 moles of Ag, reducing the concentration of Ag⁺ in the LHS to:
\[ [Ag^+]_{LHS} = 1 - 0.5 = 0.5 \text{ M} \]
- 1 F of electricity is passed through the RHS (cathode), which will deposit 1 mole of Ag, keeping the concentration of Ag⁺ in the RHS at:
\[ [Ag^+]_{RHS} = 1 \text{ M} \]

5. **Final EMF Calculation**:
- After passing the current, we recalculate the EMF using the new concentrations:
\[ E = 0 + \frac{0.0591}{1} \log \left( \frac{1}{0.5} \right) \]
- This simplifies to:
\[ E = 0 + 0.0591 \log(2) \approx 0.0591 \times 0.301 = 0.0178 \text{ volts} \]

6. **Comparison of EMF Before and After**:
- Initially, the EMF was 0 volts, and after electrolysis, the EMF is approximately 0.0178 volts.
- Thus, the EMF has increased.

### Conclusion:
The EMF of the cell after electrolysis will be **(a) Increased**.