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Q.Balance the following redox reaction in basic medium by ion-electron method: MnO4^-(aq) + I^-(aq) -> MnO2(s) + I2(s)

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2019Subjective· 4mImportance★★★★★
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Balancing by the ion-electron (half-reaction) method in basic medium gives 2MnO4^- + 4H2O + 6I^- -> 2MnO2 + 8OH^- + 3I2.

Step 1: Split into half-reactions

Reduction: MnO4^-(aq) -> MnO2(s) (Mn goes from +7 to +4, a gain of 3 electrons)

Oxidation: I^-(aq) -> I2(s) (I goes from -1 to 0, a loss of 1 electron per I atom)

Step 2: Balance atoms other than O and H, then O using H2O, then H using H+ (acidic first), then convert to basic

Reduction half-reaction:

MnO4^- -> MnO2 (Mn balanced)

Balance O by adding 2H2O to the right (MnO4^- has 4 O, MnO2 has 2 O, need 2 more O on the right from water, but water adds O to the side missing O; here MnO4^- has more O so we add H2O to the product side and H+ to reactant side following the acidic method):

MnO4^- + 4H+ -> MnO2 + 2H2O

Balance charge with electrons: MnO4^- + 4H+ + 3e- -> MnO2 + 2H2O

Convert to basic medium by adding 4OH- to both sides (to neutralise the 4H+):

MnO4^- + 4H2O + 3e- -> MnO2 + 2H2O + 4OH-

Simplify water (4H2O - 2H2O = 2H2O left):

MnO4^- + 2H2O + 3e- -> MnO2 + 4OH-

Oxidation half-reaction:

2I^- -> I2 + 2e-

…

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