The `E^(Theta)` values at 298 K corresponding to the following two reduction electrode processes are:
`(i) Cu^(+)//Cu=0.52V`
(ii) `Cu^(2+)//Cu^(+) = +0.16V`
Formulate the galvanic cell for their combination. What will be the cell potential ? Calculate the `Delta_(r )G^(Theta)` for the cell reaction `(F = 96500" C mol"^(-))`.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the half-reactions
We have two half-reactions based on the given standard reduction potentials:
1. \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \quad (E^\Theta = 0.52 \, \text{V}) \)
2. \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \quad (E^\Theta = 0.16 \, \text{V}) \)

### Step 2: Determine the anode and cathode
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