Question 23 of 31intermediate🔧 ApplyLong Answer8 marks

Balance the following redox reactions by ion–electron method: (a) MnO₄⁻ (aq) + I⁻ (aq) → MnO₂ (s) + I₂ (s) (in basic medium) (b) MnO₄⁻ (aq) + SO₂ (g) → Mn²⁺ (aq) + HSO₄⁻ (aq) (in acidic solution) (c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution) (d) Cr₂O₇²⁻ + SO₂ (g) → Cr³⁺ (aq) + SO₄²⁻ (aq) (in acidic solution)

(a)
Answer

2MnO4−(aq)+6I−(aq)+4H2O(l)→2MnO2(s)+3I2(s)+8OH−(aq)2\text{MnO}_4^-(\text{aq}) + 6\text{I}^-(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + 3\text{I}_2(\text{s}) + 8\text{OH}^-(\text{aq})

(b)
Answer

2MnO4−(aq)+5SO2(g)+2H2O(l)+H+(aq)→2Mn2+(aq)+5HSO4−(aq)2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + \text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{HSO}_4^-(\text{aq})

(c)
Answer

H2O2(aq)+2Fe2+(aq)+2H+(aq)→2Fe3+(aq)+2H2O(l)\text{H}_2\text{O}_2(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow 2\text{Fe}^{3+}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})

(d)
Answer

Cr2O72−(aq)+3SO2(g)+2H+(aq)→2Cr3+(aq)+3SO42−(aq)+H2O(l)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 3\text{SO}_2(\text{g}) + 2\text{H}^+(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l})

Explanation

To balance redox reactions by the ion-electron method, we separate the reaction into oxidation and reduction half-reactions. We balance atoms and charges in each half-reaction. For acidic medium, we balance oxygen by adding H₂O and hydrogen by adding H⁺. For basic medium, we first balance as in acidic medium, then add OH⁻ to both sides for every H⁺, combining H⁺ and OH⁻ into H₂O. Finally, we multiply the half-reactions by appropriate integers to equalize the electrons transferred and add them together.