The cell in which the following reactions occurs:
$2 \mathrm{Fe}^{3+}{ }_{(\infty)}+2 \mathrm{I}_{(a q)}^{-} \rightarrow 2 \mathrm{Fe}^{2+}{ }_{(a q)}+\mathrm{I}_{2(s)}$
has $E_{\text {cul }}^{0}=0.236 \mathrm{~V}$ at $298 \mathrm{~K}$.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Here, $n=2, E_{\text {cell }}^{\ominus}=0.236 \mathrm{~V}, \mathrm{~T}=298 \mathrm{~K}$
We know that:
$\Delta_{r} \mathrm{G}^{\ominus}=-n \mathrm{FE}_{\mathrm{cell}}^{\ominus}$
= −2 × 96487 × 0.236
= −45541.864 J mol−1
= −45.54 kJ mol−1
Again, $\Delta_{r} G^{\ominus}=-2.303 \mathrm{R} T \log K_{c}$
$\Rightarrow \log K_{c}=-\frac{\Delta_{r} G^{\ominus}}{2.303 \mathrm{R} T}$
$=-\frac{-45.54 \times 10^{3}}{2.303 \times 8.314 \times 298}$
= 7.981
∴Kc = Antilog (7.981)
= 9.57 × 107
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