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Question
Justify the following reaction as a redox reaction.
\[\ce{2Na_{(s)} + S_{(s)} -> Na2S_{(s)}}\]
Find out the oxidizing and reducing agents.
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Solution
- A redox reaction can be described as electron transfer, as shown below:
\[\ce{2Na_{(s)} + S_{(s)}->2Na^+ + S^2-}\] - Charge development suggests that each sodium atom loses one electron to form Na+ and the sulphur atom gains two electrons to form S2–. This can be represented as follows:
- When Na is oxidized to Na2S, the neutral Na atom loses electrons to form Na+ in Na2S while the elemental sulphur gains electrons and forms S2– in Na2S.
- Each of the above steps represents a half-reaction which involves electron transfer (loss or gain).
- The sum of these two half-reactions, or the overall reaction, is a redox reaction.
- An oxidizing agent is an electron acceptor, and hence, S is an oxidizing agent. A reducing agent is an electron donor, and hence, Na is a reducing agent.
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