The Gibbs energy change of the reaction (in k | vedclass

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(A) The cell reaction is: $2Cr(s) + 3Fe^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Fe(s)$.Number of electrons transferred,$n = 6$.Standard cell potential: $

Vedclass · Accepted Answer

$-150.9$

Vedclass · Answer

(A) The cell reaction is: $2Cr(s) + 3Fe^{2+}(aq) ightarrow 2Cr^{3+}(aq) + 3Fe(s)$.Number of electrons transferred,$n = 6$.Standard cell potential: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Cr^{3+}/Cr} = -0.44 - (-0.74) = +0.30 \ V$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}$.$E_{cell} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.01)^3} = 0.30 - 0.00985 \log \frac{10^{-2}}{10^{-6}} = 0.30 - 0.00985 \log(10^4) = 0.30 - 0.00985 	imes 4 = 0.30 - 0.0394 = 0.2606 \ V$.Gibbs energy change: $\Delta G = -nFE_{cell} = -6 	imes 96500 	imes 0.2606 \ J \ mol^{-1} = -150887.4 \ J \ mol^{-1} \approx -150.9 \ kJ \ mol^{-1}$.

The Gibbs energy change of the reaction (in $kJ \ mol^{-1}$) corresponding to the following cell $Cr | Cr^{3+} (0.1 \ M) || Fe^{2+} (0.01 \ M) | Fe$ is: (Given: $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \ V$,$E^{\circ}_{Fe^{2+}/Fe} = -0.44 \ V$)

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