The `E_(cell)^(0)` for the reaction `Fe+Zn^(2+)hArrZn+Fe^(3+)` is `-0.32` volt at `25^(@)C`. What will be the equilibrium concentration of `Fe^(2+)`, when a piece of iron is placed in a `1MZn^(2+)` solution?

Explanation

To find the equilibrium concentration of \( \text{Fe}^{2+} \) when a piece of iron is placed in a \( 1 \, \text{M} \, \text{Zn}^{2+} \) solution, we can follow these steps:

### Step 1: Write the given reaction and identify the standard cell potential
The reaction is:
\[ \text{Fe} + \text{Zn}^{2+} \rightleftharpoons \text{Zn} + \text{Fe}^{2+} \]
The standard cell potential \( E^\circ_{\text{cell}} \) is given as \( -0.32 \, \text{V} \).
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