Question

Match the reactions given in List I with the number of electrons lost or gained in List II
$\begin{array}{cccc}& Column-I& & Column-II\\ & Reaction& & Numberofelectronslostorgained\\ \left(P\right)& Mn(OH{)}_2+H_2{O}_2\to Mn{O}_2+2H_{2}O& \left(1\right)& 8\\ \left(Q\right)& AlC{l}_3+3K\to Al+3KCl& \left(2\right)& 2\\ \left(R\right)& 3Fe+4H_{2}O\to F{e}_3{O}_4+4H_2& \left(3\right)& 3\\ \left(S\right)& 3H_{2}S+2HN{O}_3\to 3S+2NO+4H_{2}O& \left(4\right)& 6\end{array}$

• A. $\begin{array}{cccc}P& Q& R& S\\ 2& 3& 1& 4\end{array}$
• B. $\begin{array}{cccc}P& Q& R& S\\ 3& 2& 1& 4\end{array}$
• C. $\begin{array}{cccc}P& Q& R& S\\ 2& 1& 3& 4\end{array}$
• D. $\begin{array}{cccc}P& Q& R& S\\ 1& 4& 3& 2\end{array}$

Answer 1

The correct option is A $\begin{array}{cccc}P& Q& R& S\\ 2& 3& 1& 4\end{array}$

$\left(P\right)M{n}^{2+}\to M{n}^{4+}+2e^{-}$ (Oxidation)
$2e^{-}+H_2{O}_2\to 2H_{2}O$ (Reduction)
$\left(Q\right)3K\to 3K^{+}+3e^{-}$ (Oxidation)
$3e^{-}+A{l}^{3+}\to Al$ (Reduction)
$\left(R\right)3Fe\to F{e}_3{O}_4+8e^{-}$ (Oxidation)
$3x=0$ $3x-8=0$
$3x=8$
$8e^{-}+4H_{2}O\to 4H_2$ (Reduction)

$\left(S\right)3H_{2}S\to 3S+6e^{-}$ (Oxidation )
$2+x=0$ $x=0$
$x=-2$
$6e^{-}+2N{O}_3^{-}\to 2NO$ (Reduction)