Question

the two half-reactions are given below:
$6\mathrm{e}^- + 7\mathrm{h}_2\mathrm{o} + \mathrm{cr}_2\mathrm{o}_7^{-2} \
ightarrow 2\mathrm{cr}^{+3} + 14\mathrm{oh}^-$
$\mathrm{hg} \
ightarrow \mathrm{hg}^{+2} + 2\mathrm{e}^-$
what is the coefficient for $\mathrm{h}_2\mathrm{o}$ in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in half-reactions

The first half-reaction has \(6e^-\) and the second has \(2e^-\). To balance electrons, multiply the second half-reaction by 3:
\(3\times(Hg
ightarrow Hg^{+2} + 2e^-)\) gives \(3Hg
ightarrow 3Hg^{+2} + 6e^-\)

Step2: Add the two half-reactions

Now add the first half-reaction (\(6e^- + 7H_2O + Cr_2O_7^{2-}
ightarrow 2Cr^{+3} + 14OH^-\)) and the multiplied second half-reaction (\(3Hg
ightarrow 3Hg^{+2} + 6e^-\)):
Cancel \(6e^-\) from both sides:
\(7H_2O + Cr_2O_7^{2-} + 3Hg
ightarrow 2Cr^{+3} + 14OH^- + 3Hg^{+2}\)

Answer:

7