Question

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

$2{{\mathrm{MnO}}_4}^{-}+{\mathrm{cH}}^{+}+{\mathrm{bC}}_2{{\mathrm{O}}_4}^{-}\to {\mathrm{xMn}}^{+2}+{\mathrm{zH}}_2\mathrm{O}+{\mathrm{yCO}}_2$

Answer 1

Step 1: Balancing the oxidation half:-

The oxidation part of the reaction is:

${{\mathrm{MnO}}_4}^{-}\to {\mathrm{Mn}}^{+2}$

First, we need to balance the oxygen, so we add 4 water molecules to the right side to balance it:

${{\mathrm{MnO}}_4}^{-}\to {\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O}$

Now we need to balance the hydrogen, so we add 4 hydrogens on the left side to balance it:

${{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O}$

The total charge on the left side is $+8-1=+7$, and on the right side is $+2$

So to balance this we need to add 5 electrons on the left side:

$5e^{-}+{{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O}$

Step 2: Balancing the reduction half:-

The reduction part of the reaction is:

${{\mathrm{C}}_2{\mathrm{O}}_4}^{2-}\to {\mathrm{CO}}_2$

First, we need to balance the oxygen, so we multiply the Carbon dioxide on the right side by 2 to balance it:

${{\mathrm{C}}_2{\mathrm{O}}_4}^{2-}\to 2{\mathrm{CO}}_2$

The total charge on the left side is $-2$, and on the right side is $0$

So to balance this we need to add 2 electrons on the right side:

${{\mathrm{C}}_2{\mathrm{O}}_4}^{2-}\to 2{\mathrm{CO}}_2+2{\mathrm{e}}^{-}$

Step 3: Writing the net equation

The two halves are:

$5e^{-}+{{\mathrm{MnO}}_4}^{-}+8{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{+2}+4{\mathrm{H}}_2\mathrm{O}$………Reduction

${{\mathrm{C}}_2{\mathrm{O}}_4}^{2-}\to 2{\mathrm{CO}}_2+2{\mathrm{e}}^{-}$……….Oxidation

Now to cancel out the electrons we multiply the reduction half by 2 and the oxidation half by 5

Then we add both halves to obtain our net equation:

$2{{\mathrm{MnO}}_4}^{-}+16{\mathrm{H}}^{+}+5{\mathrm{C}}_2{{\mathrm{O}}_4}^{-}\to 2{\mathrm{Mn}}^{+2}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{CO}}_2$

Now the value of is $16$

Therefore, the value of c after balancing the equation is $16$