`E_(cell)^(@)` for the given redox reaction is 2.71 V
`Mg_((s))+Cu_((0.01 M))^(2+)toMg_((0.001 M))^(2+)+Cu_((s))`
Calculate `E_(cell)` for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The given redox reaction is:
\[ \text{Mg}_{(s)} + \text{Cu}^{2+}_{(0.01 M)} \rightarrow \text{Mg}^{2+}_{(0.001 M)} + \text{Cu}_{(s)} \]

### Step 2: Identify the Standard Cell Potential
The standard cell potential \( E^\circ_{cell} \) is given as 2.71 V.
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