A button cell used in watched funcations as follwing
`Zn(s)+Ag_(2)O(s)+H_(2)O(l)hArr2Ag(s)+Zn^(2+)(aq.)+2OH^(-)(aq)`
If half cell potentials are
`Zn^(2+)(aq.)+2e^(-) rarr Zn(s), E^(@) = -0.76 V`
`Ag_(2)O(s)+H_(2)O(l)+2e^(-) rarr 2Ag(s)+2OH^(-)(aq.),, E^(@) = 0.34V`
The cell potential will be

Options

A. 1.10V
B. 0.42V
C. 0.84V
D. 1.34V

Explanation

To calculate the cell potential for the given button cell reaction, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Half-Reactions:**
The overall cell reaction is given as:
\[ \text{Zn(s)} + \text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightleftharpoons 2\text{Ag(s)} + \text{Zn}^{2+}(aq) + 2\text{OH}^-(aq) \]
From this, we can identify the half-reactions:
- Oxidation half-reaction (at the anode):
\[ \text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn(s)}, \quad E^\circ = -0.76 \, \text{V} \]
- Reduction half-reaction (at the cathode):
\[ \text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^-(aq), \quad E^\circ = 0.34 \, \text{V} \]

2. **Determine Anode and Cathode:**
- The anode is where oxidation occurs, which is the zinc half-reaction.
- The cathode is where reduction occurs, which is the silver half-reaction.

3. **Calculate the Cell Potential (E_cell):**
The cell potential (E_cell) can be calculated using the formula:
\[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) \]
\[ E_{\text{cell}} = 0.34 \, \text{V} + 0.76 \, \text{V} = 1.10 \, \text{V} \]

4. **Conclusion:**
The cell potential for the button cell is:
\[ E_{\text{cell}} = 1.10 \, \text{V} \]