• # question_answer 28) Balance the following redox reactions by ion electron method:   (a)$MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to \underset{\left( in\text{ }basic\text{ }medium \right)}{\mathop{Mn{{O}_{2}}(s)+{{I}_{2}}(s)}}\,$ (b) $MnO_{4}^{-}(aq)+S{{O}_{2}}(g)\to \underset{\left( in\text{ acidic }medium \right)}{\mathop{M{{n}^{2+}}(aq)+HSO_{4}^{-}(aq)}}\,$ (c) ${{H}_{2}}{{O}_{2}}(aq)+F{{e}^{2+}}(aq)\to F{{e}^{3+}}\underset{\left( in\text{ }acidic\text{ }medium \right)}{\mathop{(aq)+{{H}_{2}}O(l)}}\,$ (d)$C{{r}_{2}}O_{7}^{2-}+S{{O}_{2}}(g)\to C{{r}^{3+}}\underset{\left( in\text{ }acidic\text{ }medium \right)}{\mathop{(aq)+SO_{4}^{2-}(aq)}}\,$

(a) Reaction: $MnO_{4}^{-}(aq)+{{I}^{-}}(aq)\to Mn{{O}_{2}}(s)+{{I}_{2}}(s)$ 1st step: Splitting into two half reactions: (balance the elements other than hydrogen and oxygen) $MnO_{4}^{-}\to Mn{{O}_{2}}(s)$ (Reduction) $2{{I}^{-}}\to {{I}_{2}}(s)$ (Oxidation) 2nd step: Balance oxygen atoms by adding water molecule to the oxygen deficient side. $MnO_{4}^{-}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}$ 3rd step: Balance hydrogen by adding ${{H}^{+}}$ ions to the hydrogen deficient side. $MnO_{4}^{-}+4{{H}^{+}}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}$ 4th step: Balance the charge by adding electrons $MnO_{4}^{-}+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O$ $2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}$ 5th step: Equations are added in such a way that electrons are cancelled; we now get the balanced equation. $[MnO_{4}^{-}]+4{{H}^{+}}+3{{e}^{-}}\to Mn{{O}_{2}}+2{{H}_{2}}O]\times 2$ $[2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}}]\times 3$ $\frac{\frac{{}}{2MnO_{4}^{-}+6{{I}^{-}}+8{{H}^{+}}\to 2Mn{{O}_{2}}+3{{I}_{2}}+4{{H}_{2}}O}}{Balanced\text{ }equation}$ $2MnO_{4}^{-}+4{{H}_{2}}O\to 2Mn{{O}_{2}}+3\underset{(Basic\,medium)}{\mathop{{{I}_{2}}+8O{{H}^{-}}}}\,$   (b) The given equation is: $MnO_{4}^{-}(aq)+S{{O}_{2}}(g)\to M{{n}^{2+}}(aq)+\underset{\left( acidic\text{ }medium \right)}{\mathop{HSO_{4}^{-}(aq)}}\,$ Step I: Splitting into two half equations: $MnO_{4}^{-}(aq)\to M{{n}^{2+}}(aq);$ $S{{O}_{2}}(g)\to HSO_{4}^{-}(aq)$ Step II: Balancing oxygen by adding water molecules: $MnO_{4}^{-}(aq)\to \overset{2+}{\mathop{Mn}}\,(aq)+4{{H}_{2}}O$ $S{{O}_{2}}(g)+2{{H}_{2}}O\to HSO_{4}^{-}(aq)$ Step III: Balancing hydrogen by adding ${{H}^{+}}$ions: $MnO_{4}^{-}(aq)+8{{H}^{+}}\to M{{n}^{2+}}(aq)+4{{H}_{2}}O$ $S{{O}_{2}}(g)+2{{H}_{2}}O\to HSO_{4}^{-}(aq)+3{{H}^{+}}$ Step IV: Balancing the charge by adding electrons: $[MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O]\times 2$ $[S{{O}_{2}}(g)+2{{H}_{2}}O\to HSO_{4}^{-}(aq)+3{{H}^{+}}+2{{e}^{-}}]\times 5$Step V: Adding above equations, we get: $2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)+2{{H}_{2}}O+{{H}^{+}}$ $\to 2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)$ It is now balanced equation. (c)${{H}_{2}}{{O}_{2}}(aq)+F{{e}^{2+}}(aq)\to F{{e}^{3+}}(aq)+{{H}_{2}}O(l)$ Step I: Splitting of equation into two half equations: $F{{e}^{2+}}\to F{{e}^{3+}}$ ${{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O$   Step II: Oxygen and other elements are balanced hydrogen will be balanced by adding ${{H}^{+}}$ ions to the hydrogen deficient side. $F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}$ ${{H}_{2}}{{O}_{2}}+2{{H}^{+}}\to 2{{H}_{2}}O$   Step III: Charge is now balanced by adding electrons to the side where positive charge is excess or negative charge is less. $F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}$ ${{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\to 2{{H}_{2}}O$ Step IV: Add these equations in such a way that electrons are cancelled. The resultant equation will now be a balanced equation. \begin{align} & 2[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}] \\ & \frac{{{H}_{2}}{{O}_{2}}+2{{H}^{+}}+2{{e}^{-}}\to 2{{H}_{2}}O}{\frac{2F{{e}^{2+}}+{{H}_{2}}{{O}_{2}}+2{{H}^{+}}\to 2F{{e}^{3+}}+2{{H}_{2}}O}{{}}} \\ \end{align} (d) $C{{r}_{2}}O_{7}^{2-}+S{{O}_{2}}+{{H}^{+}}\to C{{r}^{3+}}+HSO_{4}^{-}+{{H}_{2}}O$ Step I: Splitting into two half reactions. $\underset{(Reduction\,half\,reaction)}{\mathop{C{{r}_{2}}O_{7}^{2-}+{{H}^{+}}\to C{{r}^{3+}}+{{H}_{2}}O}}\,$$\underset{(OXidation\,half\,reaction)}{\mathop{S{{O}_{2}}\to HSO_{4}^{-}}}\,$ Step II: Adding ${{H}^{+}}$ ions to the side deficient in hydrogen. $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\to C{{r}^{3+}}+7{{H}_{2}}O$ Step III: Adding water to the side deficient in oxygen, $S{{O}_{2}}+2{{H}_{2}}O\to HSO_{4}^{-}+3{{H}^{+}}$ Step IV: Making atoms equal on both sides, $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}\to 2C{{r}^{3+}}+7{{H}_{2}}O$ Step V: Adding electrons to the sides deficient in electrons, $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6e\to 2C{{r}^{3+}}+7{{H}_{2}}O;$ $S{{O}_{2}}+2{{H}_{2}}O\to HSO_{4}^{-}+3{{H}^{+}}+2e$ Step VI: Balancing electrons in both the half reaction, $C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6e\to 2C{{r}^{3+}}+7{{H}_{2}}O$ $[S{{O}_{2}}+2{{H}_{2}}O\to HSO_{4}^{-}+3{{H}^{+}}+2e]\times 3$Step VII: Adding both the half reactions, $C{{r}_{2}}O_{7}^{2-}+5{{H}^{+}}+3S{{O}_{2}}\to 2C{{r}^{3+}}+3HSO_{4}^{-}+{{H}_{2}}O$