The e.m.f of galvanic cell , with oxidation potentials of Zn = `+0.76 V` and that of `Cu = -0.34 V` , is ________.

Options

A. `-1.1 V`
B. `+1.1V`
C. `+0.345 V`
D. `+0.76 V`

Answer

To find the e.m.f (electromotive force) of the galvanic cell with the given oxidation potentials of zinc (Zn) and copper (Cu), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Oxidation Potentials:**
- The oxidation potential for Zn is given as \( +0.76 \, \text{V} \).
- The oxidation potential for Cu is given as \( -0.34 \, \text{V} \).

2. **Understand the Cell Reaction:**
- In a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode.
- Zinc will be oxidized (loses electrons) and copper will be reduced (gains electrons).

3. **Convert Oxidation Potential of Copper to Reduction Potential:**
- The reduction potential for copper can be calculated as the negative of its oxidation potential:
\[ E^\circ_{\text{Cu}} = -(-0.34 \, \text{V}) = +0.34 \, \text{V} \]

4. **Calculate the Standard Cell Potential (E°cell):**
- The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Here, zinc is the anode and copper is the cathode:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Zn}} = 0.34 \, \text{V} - 0.76 \, \text{V} \]

5. **Perform the Calculation:**
\[ E^\circ_{\text{cell}} = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \]

6. **Final Answer:**
- The e.m.f of the galvanic cell is \( \boxed{1.10 \, \text{V}} \).