The cell in which the following reaction occurs
`2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)++I_(2)(s)` has `E_(cell)^(0)=0.236V` at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Explanation

To solve the given problem, we need to calculate the standard Gibbs free energy change (\(\Delta G^\circ\)) and the equilibrium constant (\(K_c\)) for the cell reaction:

\[ 2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s) \]

Given data:
- Standard cell potential, \(E^\circ_{cell} = 0.236 \, V\)
- Temperature, \(T = 298 \, K\)

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