krit.club logo

Redox Reactions - Balancing of Redox Reactions

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Redox reactions involve the simultaneous occurrence of oxidation (loss of electrons/increase in oxidation state) and reduction (gain of electrons/decrease in oxidation state).

•

The Oxidation Number is the formal charge an atom would have if all bonds to atoms of different elements were 100%100\% ionic.

•

Oxidation Number Method: This method is based on the principle that the total increase in oxidation number must equal the total decrease in oxidation number.

•

Half-Reaction Method (Ion-Electron Method): The overall reaction is split into two half-reactions: the oxidation half and the reduction half. Each is balanced separately for atoms and charge before being combined.

•

In acidic medium, balance oxygen atoms by adding H2OH_2O and hydrogen atoms by adding H+H^+ ions.

•

In basic medium, balance atoms as in acidic medium, then add OH−OH^- ions to both sides equal to the number of H+H^+ ions to convert them into H2OH_2O.

📐Formulae

Total Increase in O.N.=Total Decrease in O.N.\text{Total Increase in O.N.} = \text{Total Decrease in O.N.}

Oxidation Half-Reaction: A→An++ne−\text{Oxidation Half-Reaction: } A \rightarrow A^{n+} + ne^-

Reduction Half-Reaction: B+ne−→Bn−\text{Reduction Half-Reaction: } B + ne^- \rightarrow B^{n-}

Sum of Oxidation States in a neutral molecule=0\text{Sum of Oxidation States in a neutral molecule} = 0

Sum of Oxidation States in a polyatomic ion=Charge of the ion\text{Sum of Oxidation States in a polyatomic ion} = \text{Charge of the ion}

💡Examples

Problem 1:

Balance the following redox reaction in acidic medium using the Ion-Electron Method: MnO4−(aq)+Fe2+(aq)→Mn2+(aq)+Fe3+(aq)MnO_4^-(aq) + Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + Fe^{3+}(aq)

Solution:

  1. Split into half-reactions: Oxidation: Fe2+→Fe3+Fe^{2+} \rightarrow Fe^{3+} Reduction: MnO4−→Mn2+MnO_4^- \rightarrow Mn^{2+}

  2. Balance atoms other than OO and HH: (Already balanced for FeFe and MnMn).

  3. Balance OO by adding H2OH_2O: MnO4−→Mn2++4H2OMnO_4^- \rightarrow Mn^{2+} + 4H_2O

  4. Balance HH by adding H+H^+: MnO4−+8H+→Mn2++4H2OMnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O

  5. Balance charges by adding electrons (e−e^-): Oxidation: Fe2+→Fe3++e−Fe^{2+} \rightarrow Fe^{3+} + e^- Reduction: MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

  6. Equalize electrons and add: Multiply Oxidation half by 55: 5Fe2+→5Fe3++5e−5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^- Result: MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}

Explanation:

The reduction of Permanganate involves a change in oxidation state of MnMn from +7+7 to +2+2 (5 electrons), while Iron is oxidized from +2+2 to +3+3 (1 electron). H+H^+ is added to balance oxygen in acidic medium.

Problem 2:

Balance the reaction of Permanganate ion with Bromide ion in basic medium to produce Manganese Dioxide and Bromate ion: MnO4−+Br−→MnO2+BrO3−MnO_4^- + Br^- \rightarrow MnO_2 + BrO_3^-

Solution:

  1. Oxidation numbers: MnMn in MnO4−MnO_4^- is +7+7, in MnO2MnO_2 is +4+4. BrBr in Br−Br^- is −1-1, in BrO3−BrO_3^- is +5+5.

  2. Half-reactions (acidic steps first): Red: MnO4−+4H++3e−→MnO2+2H2OMnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O Ox: Br−+3H2O→BrO3−+6H++6e−Br^- + 3H_2O \rightarrow BrO_3^- + 6H^+ + 6e^-

  3. Equalize electrons (multiply reduction by 22): 2MnO4−+8H++6e−→2MnO2+4H2O2MnO_4^- + 8H^+ + 6e^- \rightarrow 2MnO_2 + 4H_2O

  4. Add reactions: 2MnO4−+Br−+2H+→2MnO2+BrO3−+H2O2MnO_4^- + Br^- + 2H^+ \rightarrow 2MnO_2 + BrO_3^- + H_2O

  5. Convert to basic medium by adding 2OH−2OH^- to both sides: 2MnO4−+Br−+2H2O→2MnO2+BrO3−+H2O+2OH−2MnO_4^- + Br^- + 2H_2O \rightarrow 2MnO_2 + BrO_3^- + H_2O + 2OH^-

  6. Simplify: 2MnO4−(aq)+Br−(aq)+H2O(l)→2MnO2(s)+BrO3−(aq)+2OH−(aq)2MnO_4^-(aq) + Br^-(aq) + H_2O(l) \rightarrow 2MnO_2(s) + BrO_3^-(aq) + 2OH^-(aq)

Explanation:

In basic medium, H+H^+ ions are neutralized by OH−OH^- to form water, and the excess OH−OH^- remains on the appropriate side to balance the charge and hydrogen atoms.