F e ^ 3 + is reduced to F e ^ 2 + by using

F{e}^{3 +}

is reduced to

F{e}^{2 +}

by using

Options

A ${H}_{2}{O}_{2}$ in presence of $NaOH$
B $N{a}_{2}{O}_{2}$ in water
C ${H}_{2}{O}_{2}$ in presence of ${H}_{2}S{O}_{4}$
D $N{a}_{2}{O}_{2}$ in presence of ${H}_{2}S{O}_{4}$

Correct answer

A. ${H}_{2}{O}_{2}$ in presence of $NaOH$

Step-by-step solution

In basic medium (\mathrm{H}_2 \mathrm{O}_2) acts as reducing agent. (\mathrm{Na}_2 \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{O}_2) (2 \mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}-\rightarrow 2 \mathrm{Fe}^{2+}+2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2)

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