Question

$2Mn{O}_4^{-}+b{C}_2{O}_4^{2-}+c{H}^{+}\to xM{n}^{2+}+yC{O}_2+z{H}_{2}O$

If the above equation is balanced with integer coefficients, the value of c is ......
(Round off to the nearest Integer)

Answer 1

Writing the half reaction

oxidation half reaction
$Mn{O}_4^{-}\to M{n}^{2+}$

balancing oxygen
$Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$

balancing Hydrogen
$8H^{+}+Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$

balancing charge
$5e^{-}+8H^{+}+Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$

Reduction half
$C_2{O}_4^{2-}\to C{O}_2$

Balancing carbon
$C_2{O}_4^{2-}\to 2C{O}_2$

Balancing charge
$C_2{O}_4^{2-}\to 2C{O}_2+2e^{-}$

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

So c = 16