Balance this reaction using the ion-electron method in the shortest way possible:
${Fe}^{2 +} + {MnO}_{4}^{-} + H^{+} \to {Fe}^{3 +} + {Mn}^{2 +} + H_{2} O$

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Solution

Step 1: Write the equation in the ionic form:
The basic reaction with an oxidation state of each is
$+ 2 + 7 - 2 + 1 + 3 + 2 + 1 - 2 {Fe}^{2 +} (aq) + {MnO}_{4}^{-} (aq) + H^{+} (aq) \to {Fe}^{3 +} (aq) + {Mn}^{2 +} + H_{2} O (l) ⋮ ⋮ ⋮ ⋮ + 2 + 7 + 3 + 2$
The change in oxidation state should be written in ionic forms
${Fe}^{2 +} + {MnO}_{4}^{-} + H^{+} \to {Fe}^{3 +} + {Mn}^{2 +} + H_{2} O$
Step 2: Identify the atoms undergoing oxidation and reduction:
The decrease in oxidation number indicates is the reduction and increase in the oxidation state is called oxidation.
${Fe}^{2 +} + {MnO}_{4}^{-} + H^{+} \to {Fe}^{3 +} + {Mn}^{2 +} + H_{2} O ⋮ ⋮ Reduction oxidation$
Step 3: Break the equation into two halves:
$Oxidation half ⋮ Reduction half {Fe}^{2 +} \to {Fe}^{3 +} + e^{-} {MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O (+ 7 to + 2 = + 5 differ) Electron count = 1 Electron count = 5$

Step 4: Balancing the electron count and balanced equation:
Need to balance the electron count on both sides,
$5 {Fe}^{2 +} \to 5 {Fe}^{3 +} + 5 e^{-} {MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to {Mn}^{2 +} + 4 H_{2} O$
The $5 e^{-}$ gets cancelled on each part, so the net balanced equation is,
${5 {Fe}^{2 +} (aq) + {MnO}_{4}}^{-} (aq) + 8 H^{+} (aq) \to 5 {Fe}^{3 +} (aq) + {Mn}^{2 +} (aq) + 4 H_{2} O (l)$

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