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NVAMediumJEE 2024Oxidation Number & Redox Reactions

JEE Chemistry 2024 Question with Solution

2MnO4−+6I−+4H2O→3I2+2MnO2+8OH−2\text{MnO}_4^- + 6\text{I}^- + 4\text{H}_2\text{O} \rightarrow 3\text{I}_2 + 2\text{MnO}_2 + 8\text{OH}^-. If the above equation is balanced with integer coefficients, the value of zz is:

Answer

Correct answer:8

Step-by-step solution

Standard Method

Given: The redox equation is to be balanced and the coefficient zz of OH−\text{OH}^- is required.

Find: The value of zz.

Using the ion-electron method, write the half-reactions.

Reduction half reaction:

2MnO4−→2MnO22\text{MnO}_4^- \rightarrow 2\text{MnO}_2 2MnO4−+4H2O+6e−→2MnO2+8OH−2\text{MnO}_4^- + 4\text{H}_2\text{O} + 6e^- \rightarrow 2\text{MnO}_2 + 8\text{OH}^-

Oxidation half reaction:

2I−→I2+2e−2\text{I}^- \rightarrow \text{I}_2 + 2e^- 6I−→3I2+6e−6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-

Adding the oxidation and reduction half-reactions, the net reaction becomes:

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

Thus, z=8z = 8.

Therefore, the required value is 88.

Detailed Half-Reaction Method

Given: MnO4−\text{MnO}_4^- is reduced to MnO2\text{MnO}_2 and I−\text{I}^- is oxidized to I2\text{I}_2.

Find: The coefficient zz in the balanced equation.

First identify the half-reactions:

MnO4−→MnO2\text{MnO}_4^- \rightarrow \text{MnO}_2 I−→I2\text{I}^- \rightarrow \text{I}_2

Balance the reduction half-reaction in acidic medium:

MnO4−+4H++3e−→MnO2+2H2O\text{MnO}_4^- + 4\text{H}^+ + 3e^- \rightarrow \text{MnO}_2 + 2\text{H}_2\text{O}

Multiply by 22:

2MnO4−+8H++6e−→2MnO2+4H2O2\text{MnO}_4^- + 8\text{H}^+ + 6e^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O}

Balance the oxidation half-reaction:

2I−→I2+2e−2\text{I}^- \rightarrow \text{I}_2 + 2e^-

Multiply by 33:

6I−→3I2+6e−6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-

Add the two half-reactions:

2MnO4−+8H++6I−→2MnO2+4H2O+3I22\text{MnO}_4^- + 8\text{H}^+ + 6\text{I}^- \rightarrow 2\text{MnO}_2 + 4\text{H}_2\text{O} + 3\text{I}_2

Convert to basic medium by adding 8OH−8\text{OH}^- to both sides:

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

Since H++OH−=H2O\text{H}^+ + \text{OH}^- = \text{H}_2\text{O},

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

Cancel 4H2O4\text{H}_2\text{O} from both sides:

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

Comparing with the required form, the coefficient of OH−\text{OH}^- is z=8z = 8.

Therefore, the value of zz is 88.

Common mistakes

  • Balancing the reaction only in acidic medium and stopping there is incorrect because the final equation contains OH−\text{OH}^-, which indicates basic medium. After balancing in acidic medium, convert it to basic medium by adding equal amounts of OH−\text{OH}^- to both sides.

  • Not equalizing electrons before adding the half-reactions is wrong because charge cancellation will fail. Make the number of electrons same in oxidation and reduction half-reactions before combining them.

  • Forgetting to cancel common H2O\text{H}_2\text{O} molecules after converting to basic medium gives incorrect coefficients. After forming water from H+\text{H}^+ and OH−\text{OH}^-, simplify the equation by cancelling common species on both sides.

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