Calculate the equilibrium constant for the reaction at 298K.
`Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s)`
Given, `E_(Zn^(2+)//Zn)^(@) =- 0.76V` and `E_(Cu^(2+)//Cu)^(@) = +0.34 V`

Explanation

To calculate the equilibrium constant for the reaction at 298 K, we can follow these steps:

### Step 1: Write the half-reactions and identify the anode and cathode
The given reaction is:
\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

The half-reactions are:
1. Oxidation (anode):
...