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In the following redox reaction: \( \boldsymbol{Z} \boldsymbol{n}(\boldsymbol{s})+\boldsymbol{N} \boldsymbol{O}_{\boldsymbol{3}}^{-}(\boldsymbol{a} \boldsymbol{q})+\boldsymbol{H}^{+}(\boldsymbol{a} \boldsymbol{q}) \longrightarrow \) \( Z n^{2+}(a q)+N H_{4}^{+}(a q) \) \( Z n(s) \) and \( N O_{3}^{-}(a q) \) respectively are respectively:

1. oxidant and reductant
2. reductant and oxidant
3. both oxidant
4. both reductant

Correct answer: reductant and oxidant

Solution

Zn(s) goes from 0 to +2, so it is oxidized and therefore acts as the reductant. N in NO3− is reduced to NH4+, so nitrate is the oxidant.

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