A button cell used in watches functions as following
`Zn(s) + Ag _2 O(s) + H_2 O(l)`
`2 A g(s) + Zn^(2+) (aq) = 2 OH^- (aq)`
If half cell potentials are
`Zn^(2+) (aq) +2e^- rarr Zn (s) , E^@ =- 0. 76 V`
`Af_2 O(s) +H_2 O (l) +2e^- rarr`
`2Ag(s) + 2OH^- (aq) , E^@ =0. 34 V`.
The cell potential will be .

Options

A. ` 0.42 V`
B. ` 0. 84 V`
C. ` 1.10 V`
D. ` 1.34 V`

Explanation

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

### Step 1: Identify the half-reactions and their standard electrode potentials.
The half-reactions given in the problem are:
1. **Oxidation half-reaction**:
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \quad E^\circ = -0.76 \, V \]
2. **Reduction half-reaction**:
\[ Ag_2O(s) + H_2O(l) + 2e^{-} \rightarrow 2Ag(s) + 2OH^{-}(aq) \quad E^\circ = 0.34 \, V \]

### Step 2: Determine which reaction is oxidation and which is reduction.
- **Oxidation** occurs at the anode, where zinc is oxidized to zinc ions:
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-} \]
- **Reduction** occurs at the cathode, where silver ions are reduced to silver metal:
\[ Ag_2O(s) + H_2O(l) + 2e^{-} \rightarrow 2Ag(s) + 2OH^{-}(aq) \]

### Step 3: Use the formula for cell potential.
The standard cell potential \(E^\circ_{cell}\) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
In this case:
- \(E^\circ_{cathode} = 0.34 \, V\) (for the reduction of Ag)
- \(E^\circ_{anode} = -0.76 \, V\) (for the oxidation of Zn)

### Step 4: Calculate the cell potential.
Substituting the values into the formula:
\[ E^\circ_{cell} = 0.34 \, V - (-0.76 \, V) \]
\[ E^\circ_{cell} = 0.34 \, V + 0.76 \, V = 1.10 \, V \]

### Final Answer:
The cell potential \(E^\circ_{cell}\) for the button cell is \(1.10 \, V\).

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