\(3 \mathrm{Cl}_{2}+6 \mathrm{NaOH} \rightarrow \quad \mathrm{NaClO}_{3}+5 \mathrm{NaCl}+3 \mathrm{H}_{2} \mathrm{O}\)
Oxidation number of \(\mathrm{Cl}\) is 0 in \(\mathrm{Cl}_{2},-1\) in \(\mathrm{NaCl}\) and \(+5\) in \(\mathrm{NaClO}_{3}\)
$14{H^ + } + C{r_2}O_7^{2 - } + 3Ni \to 2C{r^{3 + }} + 7{H_2}O + 3N{i^{2 + }}$