Consider a Galvenic cell,
`Zn(s) |Zn^(2+) (0.1M) ||Cu^(2+) (0.1M)|Cu(s)`
by what factor, the electrolyte in anodic half cell should be diluted to increase the emf by 9 mili volt at 298K`.

Explanation

To solve the problem of how much the electrolyte in the anodic half-cell should be diluted to increase the EMF by 9 millivolts at 298 K, we can follow these steps:

### Step 1: Write the half-reactions
The half-reactions for the galvanic cell are:
- Anodic half-reaction (oxidation):
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+} + 2e^- \]
- Cathodic half-reaction (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \]

### Step 2: Write the overall cell reaction
The overall cell reaction can be written by combining the half-reactions:
\[ \text{Zn(s)} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu(s)} \]

### Step 3: Use the Nernst equation
The Nernst equation at 298 K is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]
Where \( n = 2 \) (the number of electrons transferred).

### Step 4: Set up the initial EMF equation
For the initial concentrations (both 0.1 M):
\[ E_1 = E^{\circ}_{\text{cell}} - \frac{0.059}{2} \log \left( \frac{0.1}{0.1} \right) \]
Since \(\log(1) = 0\), we have:
\[ E_1 = E^{\circ}_{\text{cell}} \]

### Step 5: Set up the EMF equation after dilution
Let’s say the concentration of \(\text{Zn}^{2+}\) is diluted by a factor of \( x \). The new concentration will be \( \frac{0.1}{x} \). The new EMF can be expressed as:
\[ E_2 = E^{\circ}_{\text{cell}} - \frac{0.059}{2} \log \left( \frac{\frac{0.1}{x}}{0.1} \right) = E^{\circ}_{\text{cell}} - \frac{0.059}{2} \log \left( \frac{1}{x} \right) \]
This simplifies to:
\[ E_2 = E^{\circ}_{\text{cell}} + \frac{0.059}{2} \log(x) \]

### Step 6: Set up the equation for the increase in EMF
According to the problem, the increase in EMF is 9 mV (or 0.009 V):
\[ E_2 - E_1 = 0.009 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \left( E^{\circ}_{\text{cell}} + \frac{0.059}{2} \log(x) \right) - E^{\circ}_{\text{cell}} = 0.009 \]
This simplifies to:
\[ \frac{0.059}{2} \log(x) = 0.009 \]

### Step 7: Solve for \( x \)
Now, solving for \( x \):
\[ \log(x) = \frac{0.009 \times 2}{0.059} \]
Calculating the right side:
\[ \log(x) = \frac{0.018}{0.059} \approx 0.305 \]
Now, converting from logarithmic form to exponential form:
\[ x \approx 10^{0.305} \approx 2 \]

### Conclusion
Thus, the electrolyte in the anodic half-cell should be diluted by a factor of **2** to increase the EMF by 9 millivolts.

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