2. Balance each of the following reactions: a) As(s) + Cl, (g) → AsCl (8) b) Cu(s) + HNO,(aq) → Cu2+ (aq) + NO(g) (in acidic solution) c) MnO; (aq) + I-(aq) → Mn?"(aq) + 1,(s) (in acidic solution)

To balance a redox reaction the first step is to know which atom is oxidated and which reduced. To know it. We need to obtain the oxidation number for all atoms. Almost always, oxidation numbers of H is +1 and for O -2.

For Cu: The oxidation number of a pure element is 0.

HNO₂: 2×O: -4, H: +1 → N: +3

NO: O: -2 → N: +2

Thus, oxidation numbers are:

Cu(s) + HNO₂ (aq) → Cu²⁺ (aq) + NO(g)

0 +3 +2 +2

The atoms which increase oxidation number are oxidated and the atoms which decrease oxidation number are reduced. Thus, Cu is oxidated and N is reduced.

The next step is separate half-reactions, thus:

HNO₂ (aq) → NO(g) Reduction

Cu(s) → Cu²⁺ (aq) Oxidation

Then, we should balance, first, elements differents of oxygen and hydrogen: They are balanced yet

Oxygens must be balanced with H₂O and hydrogens with H⁺ (Because is acidic solution):

Cu(s) → Cu²⁺ (aq) Oxidation

H⁺ + HNO₂ (aq) → NO(g) + H₂O Reduction

After this, we must balance charges with electrons:

Cu(s) → Cu²⁺ (aq) + 2e⁻ Oxidation

1e⁻ + H⁺ + HNO₂ (aq) → NO(g) + H₂O Reduction

Electron number must be the same for oxidation and reduction

Cu(s) → Cu²⁺ (aq) + 2e⁻ Oxidation

2× (1e⁻ + H⁺ + HNO₂ (aq) → NO(g) + H₂O) Reduction

Last, we should sum half-reactions and cancel out common compounds:

Cu(s) + 2e⁻ + 2H⁺ + HNO₂ (aq) → Cu²⁺ + 2e⁻ + NO(g) + H₂O

Cu(s) + 2 H⁺ + 2 HNO₂(aq) → Cu²⁺ + 2 NO(g) + 2 H₂O

c) MnO₂ (aq) + I⁻ (aq) → Mn²⁺ (aq) + I₂(s)

Oxidation number:

MnO₂ (aq) + I⁻ (aq) → Mn²⁺ (aq) + I₂(s)

+4 -1 +2 0

Half reaction:

I⁻ (aq) → I₂(s) Oxidation

MnO₂ (aq) → Mn²⁺ (aq) Reduction

Then, we should balance, first, elements differents of oxygen and hydrogen.

2 I⁻ (aq) → I₂(s) Oxidation

MnO₂ (aq) → Mn²⁺ (aq) Reduction

Oxygens must be balanced with H₂O and hydrogens with H⁺ (Because is acidic solution):

2 I⁻ (aq) → I₂(s) Oxidation

4H⁺ + MnO₂ (aq) → Mn²⁺ + 2 H₂O(aq) Reduction

Charge balance:

2 I⁻ (aq) → I₂(s) + 2e⁻ Oxidation

2e⁻ + 4H⁺ + MnO₂ (aq) → Mn²⁺ + 2 H₂O(aq) Reduction

Last, we should sum half-reactions and cancel out common compounds:

2 I⁻ + 4H⁺ + MnO₂ (aq) → I₂(s) + Mn²⁺ + 2 H₂O(aq)

I hope it helps!