Question

Balance the following equation by ion electron method.
$H{g}_2(CN{)}_2+C{e}^{4+}\to C{O}_3^{2-}+N{O}_3^{-}+Hg(OH{)}_2+C{e}^{3+}$ (basic medium).

Answer 1

Write the reaction in two half reactions,oxidation half and reduction half as:

Oxidation:$H{g}_2(CN{)}_2\to Hg(OH{)}_2+C{O}_3^{2-}+{N{O}_3}^{-}$

Reduction:${Ce}^{4+}\to C{e}^{3+}$

Balance all atoms other than $O$ and $H$.

$H{g}_2(CN{)}_2\to 2Hg(OH{)}_2+2C{O}_3^{2-}+2{N{O}_3}^{-}$

${Ce}^{4+}\to C{e}^{3+}$

Now balance the oxygen atoms by adding $H_{2}O$ molecules.

$H{g}_2(CN{)}_2+16H_{2}O\to 2Hg(OH{)}_2+2C{O}_3^{2-}+2{N{O}_3}^{-}$

${Ce}^{4+}\to C{e}^{3+}$

Now balance hydrogen atoms by adding $H^{+}$ ions.

$H{g}_2(CN{)}_2+16H_{2}O\to 2Hg(OH{)}_2+32H^{+}+2C{O}_3^{2-}+2N{O}_3^{-}$

${Ce}^{4+}\to C{e}^{3+}$

As reaction takes place in basic medium, add one $O{H}^{-}$ each side for every $H^{+}$ ion present in the reaction.Combine $O{H}^{-}$ and $H^{+}$ to form $H_{2}O$.

$H{g}_2(CN{)}_2+32O{H}^{-}\to 2Hg(OH{)}_2+16H_{2}O+2C{O}_3^{2-}+2{N{O}_3}^{-}$

${Ce}^{4+}\to C{e}^{3+}$

To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.

$H{g}_2(CN{)}_2+32O{H}^{-}\to 2Hg(OH{)}_2+16H_{2}O+2C{O}_3^{2-}+2{N{O}_3}^{-}+26e^{-}$

${Ce}^{4+}+e^{-}\to C{e}^{3+}$

Now multiply reduction half reaction by 26 and add both the reactions we will get

$H{g}_2(CN{)}_2+26C{e}^{4+}+32O{H}^{-}\to 2Hg(OH{)}_2+16H_{2}O+2C{O}_3^{2-}+2{N{O}_3}^{-}+26C{e}^{3+}$