Question

balance the reaction between so4 2 - and i - to form so3 2 - and i2 in basic solution. when you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. enter \1\ if the coefficient is \1\. so4 2 - + i - → so3 2 - + i2 (reactant, product, neither) with a coefficient of (enter 0 for neither). water appears in the balanced equation as a (enter 0 for neither). how many electrons are transferred in this reaction?

Explanation:

Step1: Write half - reactions

Oxidation: $\text{SO}_{4}^{2 - }\to\text{SO}_{3}^{2 - }$, Reduction: $\text{I}^{-}\to\text{I}_{2}$

Step2: Balance atoms other than O and H in half - reactions

Oxidation: $\text{SO}_{4}^{2 - }\to\text{SO}_{3}^{2 - }$ (already balanced for S), Reduction: $2\text{I}^{-}\to\text{I}_{2}$

Step3: Balance O atoms by adding H₂O in basic solution

Oxidation: $\text{SO}_{4}^{2 - }\to\text{SO}_{3}^{2 - }+\text{H}_{2}\text{O}$

Step4: Balance H atoms by adding H₂O and OH⁻

Oxidation: $\text{SO}_{4}^{2 - }+\text{H}_{2}\text{O}\to\text{SO}_{3}^{2 - } + 2\text{OH}^{-}$

Step5: Balance charges by adding electrons

Oxidation: $\text{SO}_{4}^{2 - }+\text{H}_{2}\text{O}+2\text{e}^{-}\to\text{SO}_{3}^{2 - } + 2\text{OH}^{-}$, Reduction: $2\text{I}^{-}\to\text{I}_{2}+2\text{e}^{-}$

Step6: Combine half - reactions

$\text{SO}_{4}^{2 - }+2\text{I}^{-}+\text{H}_{2}\text{O}\to\text{SO}_{3}^{2 - }+\text{I}_{2}+2\text{OH}^{-}$

Answer:

Coefficient of $\text{SO}_{4}^{2 - }$: 1, Coefficient of $\text{I}^{-}$: 2, Coefficient of $\text{SO}_{3}^{2 - }$: 1, Coefficient of $\text{I}_{2}$: 1, Water coefficient: 1, Electron transfer number: 2