Question

Apply the half-reaction method to balance the following redox reactions

1.  O2 + I- → I2 (in base)

2.   HNO3 + Bi2S3 → Bi(NO3)3 + NO + S (in acid)

3.  SCN- + H2O2 → NH4+ + HCO3- + HSO4- (in acid)

Answer 1

First we need to divide the equation in two halves having similar species in each equation

O2 + I- → I2

O2 à

I-àI2

Now we need to balance reaction without considering O2 or H2

It will be,

2I-àI2

Let us balance oxygen atoms using water molecules.

O2à2H2O

Now we will balance hydrogen atoms using same number of H+ atoms on another side

O2 + 4H+à2H2O

Fifth, we will use electrons to equalize the net charge on both sides.

2I-àI2 + 2e-

4e- + O2 + 4H+ à 2H2O

Sixth, we will add both equations

2I- + 4e- + O2 + 4H+ à I2 + 2e- + 2H2O

Now to get net equation we must subtract electrons

2I- + 2e- + O2 + 4H+ à I2 + 2H2O

As reaction took place in basic solution, we need to add one (OH-) for every (H+) on each side

2I- + 2e- + O2 + 4H+ + 4OH- à I2 + 2H2O + 4OH-

Now, we need to combine OH- and H+ to form H2O

4H2O + 2I- + 2e- + O2 à I2 + 2H2O + 4OH-

Now we will subtract water molecules to get final equation.

2H2O + 2I- + 2e- + O2 à I2 + 4OH-

2.

HNO3 + Bi2S3 → Bi(NO3)3 + NO + S (in acid)

Dividing in two halves

HNO3 à NO

Bi2S3 → Bi(NO3)3 + S

Balancing ions except H and O

Bi2S3 → 2Bi(NO3)3 + 3S

Balancing O by adding H2O

HNO3 à NO + 2H2O

Bi2S3 + 9H2O → 2Bi(NO3)3 + 3S

Balancing H by adding H+

3H+ + HNO3 à NO + 2H2O

Bi2S3 + 9H2O → 2Bi(NO3)3 + 3S + 18H+

Now we must balance charges by adding electrons, each electron has -1 charge.

3e- + 3H+ + HNO3 à NO + 2H2O

Bi2S3 + 9H2O → 2Bi(NO3)3 + 3S + 18H+ + 18e-

Now we must multiply one reaction with smaller number of e- to get the electron number same so we can subtract

6e- * (3e- + 3H+ + HNO3 à NO + 2H2O)

Adding both equations

18e- + 3H+ + HNO3 + Bi2S3 + 9H2O à NO + 2H2O + 2Bi(NO3)3 + 3S + 18H+ + 18e-

Solving

Bi2S3 + HNO3 + 7H2O à 2Bi(NO3)3 + NO + 3S + 15H+

3.

SCN- + H2O2 → NH4+ + HCO3- + HSO4- (in acid)

No need for balancing other ions

Balancing O by adding H2O

5H2O + SCN- + H2O2 → NH4+ + HCO3- + HSO4-

Balancing H by adding H+

5H2O + SCN- + H2O2 → NH4+ + HCO3- + HSO4- + 6H+

Now we will add electron to balance charges

-1à5+

5H2O + SCN- + H2O2 → NH4+ + HCO3- + HSO4- + 6H+ + 6e-

Multiplying left side by 6e- and cancelling 6e- from both sides.

5H2O + SCN- + H2O2 → NH4+ + HCO3- + HSO4- + 6H+