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.

Answer:

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.