Skip to content
Question of 78

Q.Balance the given equation by ion electron method in acidic medium:
Fe2+(aq) + MnO4-(aq) -> Fe3+(aq) + Mn2+(aq)

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2018Subjective· 3mImportance★★★★★
0% · 0/78 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Balancing by the ion-electron method gives MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+.

Step 1 — Identify the two half-reactions:

  • Oxidation (Fe2+ loses an electron): Fe2+ → Fe3+
  • Reduction (Mn goes from +7 in MnO4- to +2 in Mn2+): MnO4- → Mn2+

Step 2 — Balance the oxidation half-reaction (charge only, since atoms are already balanced):

Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^-

Step 3 — Balance the reduction half-reaction:

First balance O by adding H2O; then balance H by adding H+ (acidic medium); then balance charge by adding electrons.

MnO4−→Mn2++4H2OMnO_4^- \rightarrow Mn^{2+} + 4H_2O (balances the 4 oxygens using 4 H2O on the right)

MnO4−+8H+→Mn2++4H2OMnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O (balances the 8 hydrogens now needed, using 8 H+ on the left)

Check charge: left = (-1) + 8(+1) = +7; right = +2. Difference of 5, so add 5 electrons to the more positive (left) side:

MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O (now both sides = +2, balanced)

Step 4 — Equalise electrons lost and gained. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.