Question

In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half-reaction. In either method H2O(l), OH?(aq), and H+(aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditions.

Acidic solution

In acidic solution, the nitrate ion can be used to react with a number of metal ions. One such reaction is

NO3?(aq)+Sn2+(aq)?NO2(aq)+Sn4+(aq)

Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:

NO3?(aq)+Sn2+(aq)+     ????NO2(aq)+Sn4+(aq)+     ???

Part A

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms.

Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Basic solution

Potassium permanganate, KMnO4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium iodide:

MnO4?(aq)+I?(aq)?MnO2(s)+I2(aq)

Since this reaction takes place in basic solution, H2O(l) and OH?(aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: MnO4?(aq)+I?(aq)+     ????MnO2(s)+I2(aq)+     ???

Part B

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH?(aq) in the blanks where appropriate. Your answer should have six terms.

Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Answer 1

Part A)

Oxidation Half :
Sn²⁺ ----> Sn⁴⁺ + 2e^-

Reduction Half :
NO₃^- (aq) ----> NO₂(g)
Balancing equation using H₂O

NO₃^-(aq) ----> NO₂(g) + H₂O

Balancing the other side by H⁺

NO₃^- + 2H⁺ ---> NO₂(g) + H₂O

Balancing the charges:
RHS = no charge = 0
LHS = (-1) + (+2) = +1

balanced reduction half equation
NO₃^-(aq) + 2H⁺(aq) + 1e^- ---> NO₂(g) + H₂O(l)


The oxidation half equation transferrs 2 electrons and the reduction half transferrs 1.
Balancing the electrons:

(Sn²⁺ -----> Sn⁴⁺ + 2e^-)

2 x (NO₃^- + 2H⁺ + 1e^- ---> NO₂ + H₂O)

2NO₃^-(aq) + 4H⁺(aq) + 2e^- ---> 2NO₂(g) + 2H₂O(l)


Sn⁺² ----> Sn⁺⁴ + 2e^-
2NO₃^- + 4H⁺ + 2e^- ---> 2NO₂ + 2H₂O

2NO₃^- + Sn⁺²+ + 4H⁺ + 2e^- ---> 2NO₂ + Sn⁺⁴ + 2H₂O + 2e^-

The electrons cancel out
2NO₃^- + Sn⁺²+ + 4H⁺ ---> 2NO₂ + Sn⁺⁴ + 2H₂O

Part B)

Oxidation Half :

2I^- ---> I₂ + 2e^-

Reduction Half :
MnO₄^- (aq) ----> MnO₂(g)

Balancing equation using H₂O
MnO₄^- (aq) ----> MnO₂(g) + 2H₂O

Balancing the other side by H⁺
MnO₄^- (aq) + 4H⁺ ----> MnO₂(g) + 2H₂O

Balancing the charges:
RHS = no charge = 0
LHS = (-1) + (+4) = +3

balanced reduction half equation
MnO₄^- (aq) + 4H⁺ + 3e^- ----> MnO₂(g) + 2H₂O

The oxidation half equation transferrs 2 electrons and the reduction half transferrs 3.
Balancing the electrons:

3(2I^- ---> I₂ + 2e^- )

6I^- ---> 3I₂ + 6e^-

2 x (MnO₄^- (aq) + 4H⁺ + 3e^- ----> MnO₂(g) + 2H₂O)

2MnO₄^- (aq) + 8H⁺ + 6e^- ----> 2MnO₂(g) + 4H₂O


6I^- ---> 3I₂ + 6e^-
2MnO₄^- (aq) + 8H⁺ + 6e^- ----> 2MnO₂(g) + 4H₂O

2MnO₄^- (aq) + 6I^- + 8H⁺ + 6e^- ----> 2MnO₂(g) + 3I₂ + 4H₂O + 6e^-


The electrons cancel out
2MnO₄^- (aq) + 6I^- + 8H⁺ ----> 2MnO₂(g) + 3I₂ + 4H₂O

Since solution is basic, add OH^- in rhs:

2MnO₄^- (aq) + 6I^- + 4H₂O ----> 2MnO₂(g) + 3I₂ + 8OH^-