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Use the following standard electrode potentials, calculate \( \Delta G^{\circ} \) in \( \mathrm{kJ} / \mathrm{mol} \) for the indicated reaction :
\( \begin{aligned} 5 \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 5 \mathrm{Ce}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \\ \mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} & \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) ; \quad E^{\circ}=+1.51 \mathrm{~V} \\ \mathrm{Ce}^{4+}(a q)+e^{-} & \longrightarrow \mathrm{Ce}^{3+}(a q) & E^{\circ}=+1.61 \mathrm{~V} \\ 65 & \text { (b) }-24.3 & \text { (c) }-48.25 & \text { (d) }-35.2 \end{aligned} \)
(a) \( -9.65 \)
(d) \( -35.2 \)
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