Question

For the redox reaction: $Mn{O}_4^{-}+C_2{O}_4^{2-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$
the correct coefficients of the reactants for the balanced reaction are:

	$Mn{O}_4^{-}$	$C_2{O}_4^{2-}$	$H^{+}$
(a)	2	5	16
(b)	16	5	2
(c)	5	16	2
(d)	2	16	5

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

To get the balanced reaction, we write two half reactions
(i) $C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$ (Oxidation)
(ii) $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$ (Reduction) To get balanced net reaction multiply (i) by 5 and (ii) by 2 and add, we get
$2Mn{O}_4^{-}+16H^{+}5C_2{O}_4^{2-}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
From this balanced equation we get various coefficients as 2, 5 and 16 for $Mn{O}_4^{-},C_2{O}_4^{2-}$ and $H^{+}$ respectively