In an experiment, 50 ml of 0.1M solution of a salt reacted with 25 ml of 0.1M solution of Sodium sulfite. The half equation for the oxidation of sulfite ion is: SO32-(Aq) H2O→ SO42- + 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
In this redox reaction, the sulfite ion (SO32-) is oxidized to sulfate (SO42-), releasing 2 electrons. To maintain charge balance, these electrons are gained by the metal ion in the salt, reducing its oxidation number. Starting from an oxidation state of 3, the metal's oxidation number decreases to 2, as calculated by the equivalence principle: 50 ml x 0.1M x (3-n) = 25 ml x 0.1M x 2. Solving gives n = 2, confirming that option C is correct. Options A, B, and D are incorrect as they do not satisfy the stoichiometric balance of the reaction.
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