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Question
Justify that the following reaction is redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which acts as a reductant.
\[\ce{2Cu2O_{(S)} + Cu2S_{(S)}->6Cu_{(S)} + SO2_{(g)}}\]
Solution
\[\ce{2Cu2O_{(S)} + Cu2S_{(S)}->6Cu_{(S)} + SO2_{(g)}}\]
- Write an oxidation number of all the atoms of reactants and products.
- Identify the species that undergoes a change in oxidation number.
- The oxidation number of S increases from –2 to +4 and that of Cu decreases from +1 to 0. Because the oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
- The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by the gain of electrons, and therefore, Cu is an oxidising agent and itself is reduced.
Result:
- The given reaction is a redox reaction.
- Oxidant/oxidising agents (Reduced species): Cu2O/ Cu2S
- Reductant/reducing agent (Oxidised species): Cu2S
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