Solve the following


$2 \mathrm{MnO}_{4}^{-}+\mathrm{bC}_{2} \mathrm{O}_{4}^{2-}+\mathrm{cH}^{+} \rightarrow \mathrm{xMn}^{2+}+\mathrm{yCO}_{2}+\mathrm{zH}_{2} \mathrm{O}$

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



Writting the half reaction oxidation half reaction

$\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$

balancing oxygen

$\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$

balancing Hydrogen

$8 \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$

balancing charge

$5 \mathrm{e}^{-}+8 \mathrm{H}^{+}+\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}$

Reduction half

$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}$

Balancing carbon

$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{C} \mathrm{O}_{2}$

Balancing charge

$\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 2 \mathrm{CO}_{2}+2 \mathrm{e}^{-}$

Net equation

$16 \mathrm{H}^{+}+2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow 10 \mathrm{CO}_{2}+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}$

So $c=16$

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