Question

1.a)Write the equation in the ionic form : CuSO4 (aq) + Fe (s) -----> FeSO4 (aq) + Cu (s)b) Divide the above equation into oxidation and reduction half reactions

Answer 1

This is your answer.

Answer 2

Oxidation : $Fe\rightarrow Fe^{2+}+2e^-$

Reduction : $Cu^{2+}+2e^-\rightarrow Cu$

Explanation:

For the given reaction:

$CuSO_4(aq)+Fe(s)\rightarrow FeSO_4(aq)+Cu(s)$

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

$Fe\rightarrow Fe^{2+}+2e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$Cu^{2+}+2e^-\rightarrow Cu$

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