Question

Match the reactions given in column I with the number of electrons lost or gained in column II.

Answer 1

(A) ${Mn}^{2+}⟶{Mn}^{4+}+2e^{-}$ (Oxidation)
$2e^{-}+H_2{O}_2⟶2H_{2}O$ (Reduction)
Thus, 2 electrons are lost or gained in the redox reaction.
(B) $3K⟶3K^{\oplus }+3e^{-}$ (Oxidation)
$3e^{-}+{Al}^{3+}⟶Al$ (Reduction)
Thus,, 3 electrons are lost or gained in the redox reaction.
(C) $3\stackrel{0}{Fe}⟶{Fe}_3{O}_4+8e^{-}$ (Oxidation)
$3x=0$ $3x-8=0$
$3x=8$
$8e^{-}+4H_{2}O⟶4H_2$ (Reduction)
$8x-8=0$ $8x=0$
$8x=8$
Thus, 8 electrons are lost or gained in the redox reaction.
(D) $3H_{2}S⟶3S+6e^{-}$
$2+x=0$ $x=0$
$x=-2$
$6e^{-}+2N{O}_3^{\ominus }⟶2NO$
$x-6=-1$ $x-2=0$
$x=5$ $x=2$
Thus, 6 electrons are lost or gained in the redox reaction.
Thus, $A\to 2,B\to 3,C\to 1,D\to 4$ is the answer.