Redox equations are often long and difficult to balance byinspection. Luckily, e

Question

Redox equations are often long and difficult to balance byinspection. Luckily, electron transfers and oxidation numbersprovide hints for balancing such equations.
Balance the following equation:

Part B:

Explanation / Answer

Part B) Separate into oxidation and reduction half reactions (use oxidationnumbers to figure this out) Oxidation: I- --> I2 Reduction: SO42---> SO2 Balance allelements besides oxygen and hydrogen. Then balance oxygen by addingwater to the side that isdeficient Oxidation:2I- -->I2 Reduction:SO42- -->SO2 + 2H2O Balance the H by adding H+ to the sidethat is deficient Oxidation: 2I---> I2 Reduction: SO42- +4H+--> SO2 +2H2O Balance charge by adding electrons to themore positive side of theequations Oxidation:2I- -->I2 + 2e- Reduction: SO42- +4H+ + 2e- --> SO2 +2H2O Multiply a half-reaction by an integervalue so that the number of electrons is thesame (Not needed. Number of electrons are equal) Combine reactions into one. Cancel whatoccurs simultaneously on the left andright 2I- +SO42- + 4H+ --> I2 + SO2 +2H2O Reaction is 1:1 ratio between SO2 and I2,therefore one mole ofSO2 is produced when one mole ofI2 is produced.

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