Question

$F{e}^{3+}$ is reduced to $F{e}^{2+}$ by using -

• 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 $H_{2}S{O}_4$

Answer 1

Answer: (A), (B)

$F{e}^{+3}+H_2{O}_2+NaOH\to F{e}^{+2}+O_2$
$F{e}^{+3}+\left[\begin{array}{c}N{a}_2{O}_2+H_{2}O\\ \downarrow \\ NaOH+H_2{O}_2\end{array}\right]\to F{e}^{+2}+O_2$