In an experiment, 50 ml of a 0.1M solution of a metal salt reacted with 25 ml of a 0.1M solution of Sodium sulfite. The half- equation for the oxidation of sulfite ion is: (SO32-aq + H₂O → SO₄ ²⁻ + 2H⁺ + 2e⁻). If the oxidation number of the metal in the salt was +3, what would be the new oxidation number of the metal?
Correct answer: C. 2
- A. 0
- B. 1
- C. 2
- D. 4
Explanation
Equivalents of sulfite = Molarity × volume × n-factor = 0.1 × 0.025 × 2 = 0.005. Equivalents of metal salt must also be 0.005. Moles of metal salt = 0.1 × 0.05 = 0.005. The n-factor for the metal is equivalents/moles = 0.005/0.005 = 1. The change in oxidation state is 1, so the new state is +3 - 1 = +2.
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