# Question #bfe80

Oct 16, 2016

$C l {O}_{3}^{-} \left(a q\right) + \textcolor{red}{3} S {O}_{2} \left(g\right) + \textcolor{b l u e}{6} O {H}^{-} \left(a q\right) \to \textcolor{red}{3} S {O}_{4}^{2 -} \left(a q\right) + C {l}^{-} \left(a q\right) + \textcolor{red}{3} {H}_{2} O \left(l\right)$

#### Explanation:

The reaction to balance is the following:

$C l {O}_{3}^{-} \left(a q\right) + S {O}_{2} \left(g\right) \to S {O}_{4}^{2 -} \left(a q\right) + C {l}^{-} \left(a q\right)$

Step 1 : we will first split the reaction into two half equations:

Oxidation: $S {O}_{2} \to S {O}_{4}^{2 -}$

Reduction: $C l {O}_{3}^{-} \to C {l}^{-}$

Step 2 : balance the other elements other than oxygen and nitrogen.

They are balanced.

Step 3 : balance oxygen using ${H}_{2} O$:

Oxidation: $S {O}_{2} + 2 {H}_{2} O \to S {O}_{4}^{2 -}$

Reduction: $C l {O}_{3}^{-} \to C {l}^{-} + 3 {H}_{2} O$

Step 4 : balance hydrogen using ${H}^{+}$:

Oxidation: $S {O}_{2} + 2 {H}_{2} O \to S {O}_{4}^{2 -} + 4 {H}^{+}$

Reduction: $C l {O}_{3}^{-} + 6 {H}^{+} \to C {l}^{-} + 3 {H}_{2} O$

Step 5 : balance charges using ${e}^{-}$:

Oxidation: $S {O}_{2} + 2 {H}_{2} O \to S {O}_{4}^{2 -} + 4 {H}^{+} + 2 {e}^{-}$

Reduction: $C l {O}_{3}^{-} + 6 {H}^{+} + 6 {e}^{-} \to C {l}^{-} + 3 {H}_{2} O$

Step 7 : sum the two half equations while cancelling the number of electrons by multiplying each half equation by the corresponding integer:

Oxidation: $\textcolor{red}{3 \times} \left(S {O}_{2} + 2 {H}_{2} O \to S {O}_{4}^{2 -} + 4 {H}^{+} + \cancel{2 {e}^{-}}\right)$

Reduction: $\textcolor{red}{1 \times} \left(C l {O}_{3}^{-} + 6 {H}^{+} + \cancel{6 {e}^{-}} \to C {l}^{-} + 3 {H}_{2} O\right)$

RedOx: $C l {O}_{3}^{-} \left(a q\right) + \textcolor{red}{3} S {O}_{2} \left(g\right) + \textcolor{red}{3} {H}_{2} O \left(l\right) \to \textcolor{red}{3} S {O}_{4}^{2 -} \left(a q\right) + C {l}^{-} \left(a q\right) + \textcolor{red}{6} {H}^{+} \left(a q\right)$

Step 8 : cancel the ${H}^{+}$ ions by adding $O {H}^{-}$ to each side:

RedOx: $C l {O}_{3}^{-} \left(a q\right) + \textcolor{red}{3} S {O}_{2} \left(g\right) + \textcolor{red}{3} {H}_{2} O \left(l\right) + \textcolor{b l u e}{6} O {H}^{-} \left(a q\right) \to \textcolor{red}{3} S {O}_{4}^{2 -} \left(a q\right) + C {l}^{-} \left(a q\right) + {\underbrace{\textcolor{red}{6} {H}^{+} \left(a q\right) + \textcolor{b l u e}{6} O {H}^{-} \left(a q\right)}}_{6 {H}_{2} O}$

Step 9 : FINALLY, cancel the ${H}_{2} O$ molecules on both sides:

RedOx: $C l {O}_{3}^{-} \left(a q\right) + \textcolor{red}{3} S {O}_{2} \left(g\right) + \textcolor{b l u e}{6} O {H}^{-} \left(a q\right) \to \textcolor{red}{3} S {O}_{4}^{2 -} \left(a q\right) + C {l}^{-} \left(a q\right) + \textcolor{red}{3} {H}_{2} O \left(l\right)$

Here is a video that explains this topic in details:
Balancing Redox Reactions | Basic Medium.