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Q2.12 (iii):
How much charge is required for the following reduction: (iii) 1 mol of (MnO_{4}^{-}) to (Mn^{2+})?

Solution :

Step 1: Identifying the Oxidation State Change

To determine the charge required for the reduction of the permanganate ion ($MnO_{4}^{-}$) to the manganese(II) ion ($Mn^{2+}$), we must first identify the change in the oxidation state of Manganese ($Mn$).

  • In $MnO_{4}^{-}$: Let the oxidation state of $Mn$ be $x$. Since oxygen is $-2$, we have: $x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$
  • In $Mn^{2+}$: The oxidation state is given as $+2$.

The net change in the oxidation state ($\Delta OS$) is: $|7 - 2| = 5$ units [By the principle of charge conservation in redox half-reactions].

Step 2: Constructing the Balanced Half-Reaction

The reduction half-reaction can be represented as follows: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$

This balanced equation indicates that 1 mole of $MnO_{4}^{-}$ requires 5 moles of electrons ($e^{-}$) to be reduced to $Mn^{2+}$.

Step 3: Calculating Total Charge

According to Faraday’s Laws of Electrolysis, the total charge ($Q$) required is calculated by the product of the number of moles of electrons ($n$) and the Faraday constant ($F$): $Q = n \times F$

Given:

  • $n = 5 \text{ mol } e^{-}$
  • $F \approx 96500 \text{ C mol}^{-1}$ [The charge of one mole of electrons]

Step 4: Numerical Computation

$Q = 5 \text{ mol} \times 96500 \text{ C mol}^{-1}$ $Q = 482500 \text{ C}$

Visual Representation of Electron Transfer

MnO₄⁻ Gain of 5e⁻ Mn²⁺ Oxidation State: +7 Oxidation State: +2

Final Solution: The total charge required for the reduction of 1 mol of $MnO_{4}^{-}$ to $Mn^{2+}$ is 482,500 C.


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