• # question_answer 8) Write the net ionic equation for the reaction of potassium dichromate (VI) ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$with sodium sulphite$N{{a}_{2}}S{{O}_{3}}$in acid solution to give chromium (III) ion and sulphate ion.

The Skeletal equation is $C{{r}_{2}}O_{7}^{2-}(aq)+SO_{3}^{2-}(aq)\to C{{r}^{3+}}(aq)+SO_{4}^{2-}(aq)$ 1st step: Two half equations are $C{{r}_{2}}O_{7}^{2-}(aq)\to 2C{{r}^{3+}}and\,SO_{3}^{2-}(aq)\to SO_{4}^{2-}(aq)$2nd Step: Balancing the element other than hydrogen and oxygen. $C{{r}_{2}}O_{7}^{2-}(aq)\to 2C{{r}^{3+}}and\,SO_{3}^{2-}(aq)\to SO_{4}^{2-}(aq)$3rd Step: Balancing the oxygen by adding water to the oxygen deficient side. $C{{r}_{2}}O_{7}^{2-}\to 2C{{r}^{3+}}+7{{H}_{2}}O$ and$SO_{3}^{2-}(aq)+{{H}_{2}}O\to SO_{4}^{2-}(aq)$ 4th Step: Balancing hydrogen by adding ${{H}^{+}}$and change by adding electrons. $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O$ $[SO_{3}^{2-}(aq)+{{H}_{2}}O\to SO_{4}^{2-}(aq)2{{H}^{+}}+2{{e}^{-}}]3$ Adding above two equations such that the electrons are cancelled out, we get balanced chemical equation.