Question

Complete and balance the following equations, and identify the oxidizing and reducing agents. (Use the lowest possible coefficients. Include states-of-matter (s), (l), (g), (aq) similar to the given reaction.)

a.)  MnO₄^-(aq) + CH₃OH(aq) ? Mn²⁺(aq) + HCO₂H(aq) (acidic solution)

b.) H₂O₂(aq) + Cl₂O₇(aq) ? ClO₂^-(aq) + O₂(g) (basic solution)

c.) Cr₂O₇^2-(aq) + I^-(aq) ? Cr³⁺(aq) + IO₃^-(aq) (acidic solution)

d.)  As(s) + ClO₃^-(aq) ? H₃AsO₃(aq) + HClO(aq) (acidic solution)

Answer 1

a).

Given reaction can be written down in two half reactions as follows:

Oxidation half reaction:

CH₃OH + H₂O --->HCOOH + 4H⁺ + 4e-

Reduction half reaction:

MnO₄^- + 8H+ + 5e^-  ---> Mn²⁺ + 4H₂O

Therefore, CH₃OH is the reducing agent as it itself undergo oxidation(loss of electrons) and MnO₄ is the oxidsing agent as it itself undergo reduction (gain of electrons).

To balance the net equation multiply oxidation half reaction by 5 and reduction half reaction by 4 and add both the equations.

5CH₃OH + 5H₂O  ---> 5HCOOH + 20H⁺ + 20e^-

4MnO₄^- + 32H⁺ + 20e^-  ---> 4Mn²⁺ + 16H₂O

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4MnO₄^- (aq) + 5CH₃OH (aq) + 12H⁺ (aq) ---> 5HCOOH(aq) + 4Mn²⁺ (aq) + 11H₂O

b).

Given reaction can be written down in two half reactions as follows:

Oxidation half reaction:

H₂O₂ ---> O₂ + 2H⁺ + 2e^-

Reduction half reaction:

Cl₂O₇ + 6H⁺ + 8e^-  ---> 2ClO₂^- + 3H₂O

Therefore, H₂O₂ is the reducing agent as it itself undergo oxidation(loss of electrons) and Cl₂O₇ is the oxidsing agent as it itself undergo reduction (gain of electrons).

To balance the net equation multiply oxidation half reaction by 4 and add both the equations.

4H₂O₂ ---> 4O₂ + 8H⁺ + 8e^-

Cl₂O₇ + 6H⁺ + 8e^-  ---> 2ClO₂^- + 3H₂O

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4H₂O₂ + Cl₂O₇   ---> 2ClO₂^- + 4O₂ + 3H₂O + 2H⁺

Since, solution is basic add 2OH^- to both sides to get final balanced equation as follows:

4H₂O₂ (aq) + Cl₂O₇ (aq) + 2OH^- (aq)  ---> 2ClO₂^- (aq) + 4O₂ (g)+ 5H₂O

c).

Given reaction can be written down in two half reactions as follows:

Oxidation half reaction:

I^-   +   3H₂O ------> IO₃^-   + 6H⁺   + 6e^-

Reduction half reaction:

Cr₂O₇^2-   +   14H⁺   +   6e^- -----> 2Cr³⁺ + 7H₂O

Therefore, I^- is the reducing agent as it itself undergo oxidation(loss of electrons) and Cr₂O₇^2- is the oxidizing agent as it itself undergo reduction (gain of electrons).

To get the balance net equation add both the equations.

I^-   +   3H₂O ------> IO₃^-   + 6H⁺   + 6e^-

Cr₂O₇^2-   +   14H⁺   +   6e^- -----> 2Cr³⁺ + 7H₂O

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Cr₂O₇^2- (aq) + I^- (aq) +   8H⁺ (aq)   ----->  2Cr³⁺ (aq) + IO₃^- (aq) +     4H₂O

d).

Given reaction can be written down in two half reactions as follows:

Oxidation half reaction:

As   +   3H₂O ------> H₃AsO₃   + 3H⁺   + 3e^-

Reduction half reaction:

ClO₃^-   +   5H⁺   +   4e^- -----> HClO + 2H₂O

Therefore, As is the reducing agent as it itself undergo oxidation(loss of electrons) and ClO₃^- is the oxidizing agent as it itself undergo reduction (gain of electrons).

To balance the net equation multiply oxidation half reaction by 4 and reduction half reaction by 3 and add both the equations.

4As   +   12H₂O ------> 4H₃AsO₃   + 12H⁺   + 12e^-

3ClO₃^-   +   15H⁺   + 12e^- -----> 3HClO + 6H₂O

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4As (s)   +   6H₂O + 3ClO₃^- (aq) +   3H⁺ (aq) ------> 4H₃AsO₃ (aq) + 3HClO(aq)