Question

Consider the following reaction,
$xMn{O}_4^{-}+y{C}_2{O}_4^{2-}+z{H}^{+}\to xM{n}^{2+}+2yC{O}_2+\frac{z}{2}{H}_{2}O$
The values of $x,y$ and $z$ in the reaction are, respectively

• A. 5.2 and 16
• B. 2,5 and 8
• C. 2,5 and 16
• D. 5.2 and 8

Answer 1

Answer: (C)

The half equations of the reaction are $Mn{O}_4^{-}⟶M{n}^{2+}$ $C_2{O}_4^{2-}⟶C{O}_2$ The balanced balf equations are $Mn{O}_4^{-}+8H^{+}+5e^{-}⟶M{n}^{2+}+4H_{2}O$ $C_2{O}_4^{2-}⟶2C{O}_2+2e^{-}$ On equating number of electrons, we get $2Mn{O}_4^{-}+16H^{+}+10e^{-}⟶2M{n}^{2+}+8H_{2}O$ $5C_2{O}_4^{2-}⟶10C{O}_2+10e^{-}$ On adding both the equations, we get $2Mn{O}_4^{-}+5C_2{O}_4^{-}+16H^{+}⟶2M{n}^{2+}+2\times 5C{O}_2+\frac{16}{2}{H}_{2}O$ Thus $x,y$ and $z$ are $2,5$ and $16$ respectively.