In the button cells widely used in watCHMes and other devices the following reaction takes place :
`Zn(s)+Ag_(2)O(s)+H_(2)O(l) rarr Zn^(2+)(aq)+2Ag(s)+2OH^(c-)(aq)`
Determine `Delta_(r)G^(c-)` and `E^(c-)` for the reaction.

Explanation

To determine \(\Delta_rG^\circ\) and \(E^\circ\) for the given reaction, we will follow these steps:

### Step 1: Identify the half-reactions
The overall reaction is:
\[ \text{Zn}(s) + \text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s) + 2\text{OH}^-(aq) \]

We can break this down into two half-reactions:

1. **Anode (oxidation)**:
\[ \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \]

2. **Cathode (reduction)**:
\[ \text{Ag}_2\text{O}(s) + 2e^- \rightarrow 2\text{Ag}(s) + 2\text{OH}^-(aq) \]

### Step 2: Determine the standard electrode potentials
From standard electrode potential tables, we find:

- For the reduction of silver oxide:
\[ E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} = +0.344 \text{ V} \]

- For the oxidation of zinc:
\[ E^\circ_{\text{Zn}/\text{Zn}^{2+}} = -0.76 \text{ V} \]

### Step 3: Calculate the standard cell potential \(E^\circ\)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 0.344 \text{ V} - (-0.76 \text{ V}) = 0.344 \text{ V} + 0.76 \text{ V} = 1.104 \text{ V} \]

### Step 4: Determine the number of electrons transferred (n)
From the half-reaction, we see that 2 moles of electrons are transferred in the reaction:
\[ n = 2 \]

### Step 5: Calculate \(\Delta_rG^\circ\)
Using the Gibbs free energy equation:
\[ \Delta_rG^\circ = -nFE^\circ_{\text{cell}} \]
Where \(F\) (Faraday's constant) is approximately \(96500 \text{ C/mol}\). Substituting the values:
\[ \Delta_rG^\circ = -2 \times 96500 \text{ C/mol} \times 1.104 \text{ V} \]
Calculating this gives:
\[ \Delta_rG^\circ = -2 \times 96500 \times 1.104 = -213000 \text{ J} \approx -2.13 \times 10^5 \text{ J} \]

### Final Results
- \(\Delta_rG^\circ \approx -2.13 \times 10^5 \text{ J}\)
- \(E^\circ \approx 1.104 \text{ V}\)