Identify the correctly balanced redox reaction using ion-electron method.
$F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}$

10 F e S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 2 M n S O_{4} + 8 H_{2} O + K_{2} S O_{4}

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2 F e S O_{4} + 10 K M n O_{4} + 4 H_{2} S O_{4} \to 2 F e_{2} (S O_{4})_{3} + 5 M n S O_{4} + 8 H_{2} O + 10 K_{2} S O_{4}

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8 F e S O_{4} + 10 K M n O_{4} + 2 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 4 M n S O_{4} + H_{2} O + 2 K_{2} S O_{4}

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10 F e S O_{4} + 8 K M n O_{4} + 2 H_{2} S O_{4} \to 4 F e_{2} (S O_{4})_{3} + 2 M n S O_{4} + 5 H_{2} O + K_{2} S O_{4}

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Solution

The correct option is A $10 F e S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 2 M n S O_{4} + 8 H_{2} O + K_{2} S O_{4}$
Step-1 : Assign the oxidation number to each element present in the reaction.
$+ 2 + 6 - 2 F e S O_{4} + + 1 + 7 - 2 K M n O_{4} + + 1 + 6 - 2 H_{2} S O_{4} \to + 3 + 6 - 2 F e_{2} (S O_{4})_{3} + + 2 + 6 - 2 M n S O_{4} + + 1 - 2 H_{2} O$

Step 2. Remove all the spectator ions and convert the reaction into its ionic form.
Spectator ions - The ions whose oxidation numbers do not change during the reaction

$F e^{2 +} + + 7 M n O_{4}^{-} \to F e^{3 +} + M n^{2 +}$

Step 3 : Split the reaction into two halves i.e. the oxidation half and the reduction half
Determine the oxidation states of the atoms and balance the reaction atom-wise and then electrons-wise.
$\begin{matrix} F e^{2 +} o x i d a t i o n - ---- \to F e^{3 +} \end{matrix} ∣ ∣ M n O_{4}^{-} R e d u c t i o n - ----- \to M n^{2 +}$

Step 4: Balance the atom other than oxygen and hydrogen atom in both half reactions
$\begin{matrix} F e^{2 +} \to F e^{3 +} \end{matrix} | M n O_{4}^{-} \to M n^{2 +}$

Fe and Mn atoms are balanced on both side.

Step 5: Now balance O and H atom by $H_{2} O a n d H^{+}$ respectively by the following way: For one excess oxygen atom, add one $H_{2} O$ on the other side and two $H^{+}$ on the same side.
$F e^{2 +} \to F e^{3 +}$ (no oxygen atom) ...(i)
$8 H^{+} + M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$ ...(ii)

Step 6: Equation (i) and (ii) are balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side.
$F e^{2 +} o x i d a t i o n - ---- \to F e^{3 +} + e^{-}$ ...(1)
$5 e^{-} + 8 H^{+} + M n O_{4}^{-} R e d u c t i o n - ----- \to M n^{2 +} + 4 H_{2} O$ ...(2)

Step 7: The number of electrons gained and lost in each half reaction are equalised by multiplying both the half reactions with a suitable factor and finally the half reactions are added to give the overall balanced reaction.
Here, we multiply equation (1) by 5 and (2) by 1 and add them:
$F e^{2 +} \to F e^{3 +} + e^{-}$ ...(1) $\times$ 5

$5 e^{-} + 8 H^{+} + M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$ ...(2) $\times$ 1
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$5 F e^{2 +} + 8 H^{+} + M n O_{4}^{-} \to 5 F e^{2 +} + M n^{2 +} + 4 H_{2} O$
(Here, at this stage, you will get balanced redox reaction in Ionic form)

Step IX: Now convert the Ionic reaction into molecular form by adding the elements or species, which are removed in step (2).
Now, by some manipulation, you will get:

$5 F e S O_{4} + K M n O_{4} + 4 H_{2} S O_{4} \to \frac{5}{2} F e_{2} (S O_{4})_{3} + M n S O_{4} + 4 H_{2} O + \frac{1}{2} K_{2} S O_{4}$
or $10 F e S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 5 F e_{2} (S O_{4})_{3} + 2 M n S O_{4} + 8 H_{2} O + K_{2} S O_{4} .$

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