Question

For the redox reaction: $MnO^{-}_{4}+C_{2}O^{2-}_{4}+H^{+} \to Mn^{2+}+CO_{2}+H_{2}O$
the correct coefficients of the reactants for the balanced reaction are:

	$MnO^{-}_{4}$	$C_{2}O^{2-}_{4}$	$H^{+}$
(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) $C_{2}O^{2-}_{4} \to 2CO_{2}+2e^{-}$ (Oxidation)
(ii) $MnO^{-}_{4}+8H^{+}+5e^{-} \to Mn^{2+}+4H_{2}O$ (Reduction) To get balanced net reaction multiply (i) by 5 and (ii) by 2 and add, we get
$2MnO^{-}_{4}+16H^{+}\,5C_{2}O^{2-}_{4} \to 2Mn^{2+}+10CO_{2}+8H_{2}O$
From this balanced equation we get various coefficients as 2, 5 and 16 for $MnO^{-}_{4}, C_{2}O^{2-}_{4}$ and $H^{+}$ respectively