Exercise · Q20
Q.Answer the following question.
Justify the following reaction as redox reaction.
2 Na (s) + S (s) → Na2S (s)
Find out the oxidizing and reducing agents.
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Start your 14-day free trial to unlock the full solution →Step 1. Assign oxidation numbers. Both Na and S are free elements on the left, so both are at ON=0 (rule 1). In the ionic product Na2S, S is a monoatomic anion so ON(S) = its charge = -2 (rule 2); for the neutral compound, 2xON(Na) + (-2) = 0, so ON(Na) = +1.
Step 2. Identify the changes. Na rises from 0 to +1 (a loss of 1 electron per atom, x2 atoms = 2 electrons lost total) -- this is oxidation. S falls from 0 to -2 (a gain of 2 electrons) -- this is reduction. The 2 electrons lost by the two Na atoms exactly match the 2 electrons gained by the one S atom, confirming the equation …
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