MHT CET Chemistry question
In an experiment 50 mL of 0.1 solution of a salt reacted with 25 mL of 0.1 solution of sodium sulphite. The half equation for the oxidation of sulphite ion is: \mathrm{SO_3^{2-}(aq)} + H_2O(l) \longrightarrow \mathrm{SO_4^{2-}(aq)} + 2H^+(aq) + 2e^- If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal?
Answer: (C)
Solution
Metal salt = 5 mmol and sulphite = 2.5 mmol, which releases 5 mmol of electrons, so each metal ion gains 1 electron and +3
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