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.
Standard electrode potential for \( \text{Sn}^{4+}/\text{Sn}^{2+} \) couple is +0.15 V and that for the \( \text{Cr}^{3+}/\text{Cr} \) couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be:
To calculate the cell potential (\( E^\circ_{\text{cell}} \)), we use the standard electrode potentials of the given redox couples.
Given data:
\( E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15V \)
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74V \)