Question

For the redox reaction,
$x{C}_2{O}_4^{-2}+yMn{O}_4^{-}+z{H}^{+}\to yM{n}^{+2}+\frac{z}{2}{H}_{2}O+2xC{O}_2$
x, y and z respectively are:

• A. 2, 5, 16
• B. 5, 2, 10
• C. 5, 2, 16
• D. 5, 5, 16

Answer 1

The correct option is C

5, 2, 16

Let's begin by examining the change in the oxidation states. In the oxalate anion, $C$ is in +3 oxidation state, which becomes +4 in $C{O}_2$. So it gets oxidized. Also, the medium is acidic and the manganate, where $Mn$ is in +7 oxidation state, gets reduced to Mn(II) in $M{n}^{2+}$. Let us write the half reactions:

$[\stackrel{+3}{C_2}{O}_4^{-2}⟶2{\stackrel{+4}{CO}}_2+2e^{-}]\times 5$

$[5e^{-}+8H^{+}+\stackrel{+7}{Mn}{O}_4⟶{\stackrel{+2}{Mn}}^{+2}+4H_{2}O]\times 2$

Hence, the net reaction is:

$5C_2{O}_4^{-2}+2Mn{O}_4^{-}+16H^{+}\to 2M{n}^{+2}+8H_{2}O+10C{O}_2$

So, x =5; y = 2 and z = 16