Question

Problem 2.

Complete and balance the following equations for the reactions occurring in basic solution:

a. MnO4-(aq)+C2O42-(aq)→MnO2(s)+CO2(g)

b. NO2-(aq) + Al(s) → NH3(g) + Al(OH)4-(aq)
Answer:
a. 4H2O + 2MnO4- + 3C2O42- → 2MnO2 + 8OH- + 6CO2

b. 5H2O + NO2- + OH- + 2Al → NH3+ 2Al(OH)4-

Answer 1

Consider given reaction, MnO₄^- (aq) + C₂O₄^2- (aq)→MnO₂ (s) + CO₂ (g)

From above reaction , we get oxidation & reduction half equations as shown below.

Oxidation : C₂O₄^2- (aq) $\rightarrow$ CO₂ (g)

Reduction : MnO ₄^- (aq)  $\rightarrow$ MnO ₂ (s)  

Now, balance the atoms other than H & O in both half equations.

Oxidation : C₂O₄^2- (aq) $\rightarrow$ 2 CO₂ (g)

Reduction : MnO ₄^- (aq)  $\rightarrow$ MnO ₂ (s)  

Balance half equations for O atoms by adding H₂O to the side of half equations with less number of O atoms.

Oxidation : C₂O₄^2- (aq) $\rightarrow$ 2 CO₂ (g)

Reduction : MnO ₄^- (aq)  $\rightarrow$ MnO ₂ (s) + 2 H₂O (l)

Balance half equations for H atoms by adding H⁺to the side of half equations with less number of H atoms.

Oxidation : C₂O₄^2- (aq) $\rightarrow$ 2 CO₂ (g)

Reduction : MnO ₄^- (aq) + 4 H ⁺ (aq) $\rightarrow$ MnO ₂ (s) + 2 H₂O (l)

Balance above equations for charge by adding electrons.

Oxidation : C₂O₄^2- (aq) $\rightarrow$ 2 CO₂ (g) + 2 e ^-

Reduction : MnO ₄^- (aq) + 4 H ⁺ (aq) + 3 e ^-  $\rightarrow$ MnO ₂ (s) + 2 H₂O (l)  

Now, we can make number of electrons in two half equations equal by multiplying oxidation equation by 3 & reduction equation by 2.

Oxidation : 3 C₂O₄^2- (aq) $\rightarrow$ 6 CO₂ (g) + 6 e ^-

Reduction : 2 MnO ₄^- (aq) + 8 H ⁺ (aq) + 6 e ^-  $\rightarrow$2 MnO ₂ (s) + 4 H₂O (l)  

Add two half equations to get overall equation.

3 C₂O₄^2- (aq) $\rightarrow$ 6 CO₂ (g) + 6 e ^-

2 MnO ₄^- (aq) + 8 H ⁺ (aq) + 6 e ^-  $\rightarrow$2 MnO ₂ (s) + 4 H₂O (l)  

Overall : 3 C₂O₄^2- (aq) + 2 MnO ₄^- (aq) + 8 H ⁺ (aq) $\rightarrow$   6 CO₂ (g) + 2 MnO ₂ (s) + 4 H₂O (l)  

Above reaction takes place in basic medium.Hence add OH ^- ions equal to H⁺ ions on both sides of above equation.

3 C₂O₄^2- (aq) + 2 MnO ₄^- (aq) + 8 H ⁺ (aq) + 8 OH ^- (aq)$\rightarrow$  6 CO₂ (g) + 2 MnO ₂ (s) + 4 H₂O (l) + 8 OH ^- (aq)

We have, H ⁺ (aq) + OH ^- (aq) $\rightarrow$ H₂O (l)

$\therefore$ 3 C₂O₄^2- (aq) + 2 MnO ₄^- (aq) + 8 H₂O (l)  $\rightarrow$  6 CO₂ (g) + 2 MnO ₂ (s) + 4 H₂O (l) + 8 OH ^- (aq)

On cancelling H₂O (l) we get our final redox equation.

3 C₂O₄^2- (aq) + 2 MnO ₄^- (aq) + 4 H₂O (l)  $\rightarrow$  6 CO₂ (g) + 2 MnO ₂ (s) + 8 OH ^- (aq)

b) Consider a given reaction, NO ₂^- (aq) + Al (s) $\rightarrow$ NH ₃ (g) + Al(OH) ₄ - (aq)

From above reaction , we get oxidation & reduction half equations as shown below.

Oxidation : Al (s) $\rightarrow$ Al(OH) ₄ - (aq)

Reduction : NO ₂^- (aq)  $\rightarrow$ NH ₃ (g)

Balance half equations for O atoms by adding H₂O to the side of half equations with less number of O atoms.

Oxidation : Al (s) + 4 H₂O (l)  $\rightarrow$ Al(OH) ₄^- (aq)

Reduction : NO ₂^- (aq)  $\rightarrow$ NH ₃ (g) + 2 H₂O (l)

Balance half equations for H atoms by adding H⁺to the side of half equations with less number of H atoms.

Oxidation : Al (s) + 4 H₂O (l)  $\rightarrow$ Al(OH) ₄ - (aq) + 4 H ⁺ (aq)

Reduction : NO ₂^- (aq) + 7 H ⁺ (aq)  $\rightarrow$ NH ₃ (g) + 2 H₂O (l)

Balance above equations for charge by adding electrons.

Oxidation : Al (s) + 4 H₂O (l)  $\rightarrow$ Al(OH) ₄^- (aq) + 4 H ⁺ (aq) +3 e ^-

Reduction : NO ₂^- (aq) + 7 H ⁺ (aq) + 6  e ^-  $\rightarrow$ NH ₃ (g) + 2 H₂O (l)

Now, we can make number of electrons in two half equations equal by multiplying oxidation equation by 2

Oxidation : 2 Al (s) + 8 H₂O (l)  $\rightarrow$ 2 Al(OH) ₄^- (aq) + 8 H ⁺ (aq) + 6 e ^-

Reduction : NO ₂^- (aq) + 7 H ⁺ (aq) + 6  e ^-  $\rightarrow$ NH ₃ (g) + 2 H₂O (l)

Add two half equations to get overall equation.

2 Al (s) + 8 H₂O (l)  $\rightarrow$ 2 Al(OH) ₄^- (aq) + 8 H ⁺ (aq) + 6 e ^-

NO ₂^- (aq) + 7 H ⁺ (aq) + 6  e ^-​​​​​​​  $\rightarrow$ NH ₃ (g) + 2 H₂O (l)

OVERALL : 2 Al (s) + NO ₂^- (aq) + 6 H₂O (l)  $\rightarrow$ 2 Al(OH) ₄^- (aq) + NH ₃ (g) + H ⁺ (aq)

Above reaction takes place in basic medium.Hence add OH ^- ions equal to H⁺ ions on both sides of above equation.

2 Al (s) + NO ₂^- (aq) + 6 H₂O (l)+ OH ^- (aq) $\rightarrow$ 2 Al(OH) ₄^- (aq) + NH ₃ (g) + H ⁺ (aq)​​​​​​​+ OH ^- (aq)

We have, H ⁺ (aq) + OH ^- (aq) $\rightarrow$ H₂O (l)

$\therefore$ 2 Al (s) + NO ₂^- (aq) + 6 H₂O (l)+ OH ^- (aq) $\rightarrow$ 2 Al(OH) ₄^- (aq) + NH ₃ (g) + H₂O (l)

$\therefore$2 Al (s) + NO ₂^- (aq) + 5 H₂O (l)+ OH ^- (aq) $\rightarrow$ 2 Al(OH) ₄^- (aq) + NH ₃ (g)

This is the balanced redox equation in basic medium.