What amount of electric charge is required for the reduction of 1 mole of MnO$_4^{2-}$ into Mn$^{2+}$?
To reduce MnO$_4^{2-}$ to Mn$^{2+}$, we need to consider the change in oxidation state of Mn. In MnO$_4^{2-}$, Mn is in the +2 oxidation state, and in Mn$^{2+}$, Mn is also in the +2 state. Thus, no change in oxidation state occurs. However, the reduction process involves the addition of 2 electrons per ion, and for 1 mole of MnO$_4^{2-}$, 4 moles of electrons are required.
This corresponds to a charge of 4 Faradays.
(a) Calculate the standard Gibbs energy (\(\Delta G^\circ\)) of the following reaction at 25°C:
\(\text{Au(s) + Ca\(^{2+}\)(1M) $\rightarrow$ Au\(^{3+}\)(1M) + Ca(s)} \)
\(\text{E\(^\circ_{\text{Au}^{3+}/\text{Au}} = +1.5 V, E\)\(^\circ_{\text{Ca}^{2+}/\text{Ca}} = -2.87 V\)}\)
\(\text{1 F} = 96500 C mol^{-1}\)