Question

The reducing agent in the reaction, $C_2{O}_4^{2-}+Mn{O}_4^{-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$ is:

• A. $C_2{O}_4^{2-}$
• B. $Mn{O}_4^{-}$
• C. $M{n}^{2+}$
• D. $H^{+}$

Answer 1

Answer: (A)

${\left[C_2^{+3}{O}_4\right]}^{2-}+{\left[M_n^{+7}{O}_4\right]}^{-1}$
$M_n^{+2}+C{O}_2^{+4}+H_{2}O$
Reducing agent is that whose oxidation number increases.