The e.m.f of the following galvanic cells:

Options

A. `Zn"|"z^(2+)(1M)"||"Cu^(2+)(1M)"|"Cu`
B. `Zn"|"z^(2+)(0.1M)"||"Cu^(2+)(1M)"|"Cu`
C. `Zn"|"z^(2+)(1M)"||"Cu^(2+)(0.1M)"|"Cu`
D. `Zn"|"z^(2+)(0.1M)"||"Cu^(2+)(0.1M)"|"Cu`

Answer

To find the e.m.f. (electromotive force) of the given galvanic cells, we will use the Nernst equation. The Nernst equation is given by:

\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{2.303RT}{nF} \log \frac{[\text{reduced species}]}{[\text{oxidized species}]} \]

Where:
- \(E_{\text{cell}}\) is the cell potential.
- \(E^{\circ}_{\text{cell}}\) is the standard cell potential.
- \(R\) is the universal gas constant (8.314 J/(mol·K)).
- \(T\) is the temperature in Kelvin.
- \(n\) is the number of moles of electrons transferred in the reaction.
- \(F\) is Faraday's constant (96485 C/mol).

### Step-by-step solution:

**1. First Cell:**
- Reaction: \( \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)
- Concentrations: \([\text{Cu}^{2+}] = 1 \, \text{M}, \, [\text{Zn}^{2+}] = 1 \, \text{M}\)
- Applying Nernst equation:

\[ E_1 = E^{\circ}_{\text{cell}} - \frac{2.303RT}{2F} \log \frac{1}{1} \]

- Since \(\log(1) = 0\):

\[ E_1 = E^{\circ}_{\text{cell}} \]

**2. Second Cell:**
- Reaction: \( \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)
- Concentrations: \([\text{Cu}^{2+}] = 1 \, \text{M}, \, [\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- Applying Nernst equation:

\[ E_2 = E^{\circ}_{\text{cell}} - \frac{2.303RT}{2F} \log \frac{0.1}{1} \]

- \(\log(0.1) = -1\):

\[ E_2 = E^{\circ}_{\text{cell}} + \frac{2.303RT}{2F} \]

**3. Third Cell:**
- Reaction: \( \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)
- Concentrations: \([\text{Cu}^{2+}] = 1 \, \text{M}, \, [\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- Applying Nernst equation:

\[ E_3 = E^{\circ}_{\text{cell}} - \frac{2.303RT}{2F} \log \frac{0.1}{1} \]

- \(\log(0.1) = -1\):

\[ E_3 = E^{\circ}_{\text{cell}} + \frac{2.303RT}{2F} \]

**4. Fourth Cell:**
- Reaction: \( \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \)
- Concentrations: \([\text{Cu}^{2+}] = 0.1 \, \text{M}, \, [\text{Zn}^{2+}] = 0.1 \, \text{M}\)
- Applying Nernst equation:

\[ E_4 = E^{\circ}_{\text{cell}} - \frac{2.303RT}{2F} \log \frac{0.1}{0.1} \]

- Since \(\log(1) = 0\):

\[ E_4 = E^{\circ}_{\text{cell}} \]

### Summary of Results:
- \(E_1 = E^{\circ}_{\text{cell}}\)
- \(E_2 = E^{\circ}_{\text{cell}} + \frac{2.303RT}{2F}\)
- \(E_3 = E^{\circ}_{\text{cell}} + \frac{2.303RT}{2F}\)
- \(E_4 = E^{\circ}_{\text{cell}}\)