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Q.Balance the following equation in basic medium by ion-electron method: MnO4-(aq) + I-(aq) → MnO2(s) + I2(s)

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2023Subjective· 3mImportance★★★★★
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Splitting into reduction and oxidation half-reactions, balancing atoms/charge with OH⁻/H2O (basic medium), then equalising electrons gives the fully balanced equation.

Step 1 — Identify half-reactions:

Reduction (Mn: +7 → +4): MnO4−→MnO2MnO_4^- \rightarrow MnO_2

Oxidation (I: −1 → 0): I−→I2I^- \rightarrow I_2

Step 2 — Balance each half-reaction (basic medium, ion-electron method):

Reduction:

Balance O using H2O, balance H using OH⁻ (basic medium), balance charge using electrons.

MnO4−→MnO2MnO_4^- \rightarrow MnO_2 (O: 4 → 2, need 2 H2O on product side to balance the O difference as OH-, or add H2O/OH- appropriately)

MnO4−+2H2O+3e−→MnO2+4OH−MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-

Check: O: left = 4+2=6, right = 2+4=6 ✓. H: left=4, right=4 ✓. Charge: left = −1+(−3)=−4, right=−4 ✓.

Oxidation:

2I−→I2+2e−2I^- \rightarrow I_2 + 2e^-

Step 3 — Equalise electrons (LCM of 3 and 2 is 6):

Multiply reduction half-reaction by 2:

2MnO4−+4H2O+6e−→2MnO2+8OH−2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-

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