For the cells in opposition,
Zn(s) | ZnCl_(2)(sol).|AgCl(s)|Ag|AgCl(s)|
`C-(1) = 0.02 M` , `ZnCl_(2)(sol)`| Zn(s)
`C_(2) = 0.5M`
Find out the emf (in millivolt) of the resultant cell. (take log 2 `=0.3,(RT)/F` at 298 K = 0.060)

Explanation

To find the emf (electromotive force) of the given cell, we will use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Identify the half-reactions
For the cell given:
- At the anode (oxidation):
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]
- At the cathode (reduction):
\[ \text{Ag}^{+}(aq) + e^- \rightarrow \text{Ag}(s) \]

### Step 2: Write the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \(E^{\circ}_{\text{cell}}\) is the standard cell potential (which we will assume to be 0 for this calculation).
- \(R\) is the universal gas constant (8.314 J/(mol·K)).
- \(T\) is the temperature in Kelvin (298 K).
- \(n\) is the number of moles of electrons transferred (2 for this reaction).
- \(F\) is Faraday's constant (96485 C/mol).
- \(Q\) is the reaction quotient.

### Step 3: Calculate the reaction quotient \(Q\)
For our cell, the reaction quotient \(Q\) is given by:
\[ Q = \frac{[\text{Zn}^{2+}]_{\text{anode}}}{[\text{Zn}^{2+}]_{\text{cathode}}} \]
Given:
- \([\text{Zn}^{2+}]_{\text{anode}} = 0.02 \, M\)
- \([\text{Zn}^{2+}]_{\text{cathode}} = 0.5 \, M\)

So,
\[ Q = \frac{0.02}{0.5} = 0.04 \]

### Step 4: Substitute values into the Nernst equation
Using the provided value for \(\frac{RT}{F} = 0.060\) and \(n = 2\):
\[ E_{\text{cell}} = 0 - \frac{0.060}{2} \log(0.04) \]

### Step 5: Calculate \(\log(0.04)\)
Using the approximation \(\log(0.04) = \log(4 \times 10^{-2}) = \log(4) + \log(10^{-2})\):
\[ \log(4) \approx 0.6 \quad \text{(since log 2 = 0.3)} \]
\[ \log(0.04) \approx 0.6 - 2 = -1.4 \]

### Step 6: Substitute \(\log(0.04)\) back into the equation
\[ E_{\text{cell}} = -\frac{0.060}{2} \times (-1.4) \]
\[ E_{\text{cell}} = \frac{0.060 \times 1.4}{2} = \frac{0.084}{2} = 0.042 \, V \]

### Step 7: Convert to millivolts
\[ E_{\text{cell}} = 0.042 \, V \times 1000 = 42 \, mV \]

### Final Answer
The emf of the resultant cell is:
\[ \text{EMF} = 42 \, \text{mV} \]