Question

Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations: (a) Mg(s)+NiCl2(aq)⟶MgCl2(aq)+Ni(s) (b) PCl3(l)+Cl2(g)⟶PCl5(s) (c) C2H4(g)+3O2(g)⟶2CO2(g)+2H2O(g) (d) Zn(s)+H2SO4(aq)⟶ZnSO4(aq)+H2(g) (e) 2K2S2O3(s)+I2(s)⟶K2S4O6(s)+2KI(s) (f) 3Cu(s)+8HNO3(aq)⟶3Cu(NO3)2(aq)+2NO(g)+4H2O(l)

Answer 1

Step-by-step explanation:

Step 1. Assign oxidation numbers to all

elements

PbS(s) + O2

(g) PbO(s) + SO2

(g)

• Step 2. Identify oxidized and reduced species

– PbS was oxidized (O.N. of S: -2 -> +4)

– O2 was reduced (O.N. of O: 0 -> -2)

• Step 3. Compute e-

lost and e- gained

– In the oxidation: 6e- were lost from S

– In the reduction: 2e- were gained by each O

Answer 2

An oxidizing agent, gains electrons and is reduced in a chemical reaction. It is usually in the highest oxidation state as it gains (accepts) electrons and is reduced.

A reducing agent loses electrons and is oxidized in a chemical reaction. A reducing agent is typically in one of its lower possible oxidation states, because it loses electrons (electron donor).

Having noted this, the following are the oxidation states of the substances in each equation.

Step-by-step explanation:

Mg(s )+ NiCl2(aq)⟶MgCl2(aq)+ Ni(s) :

• Mg is oxidized because it has lost electrons (oxidation number 0 to +2);
• Ni is reduced as it has gained electrons (oxidation number 2+ to 0)

PCl3(l )+ Cl2(g)⟶ PCl5(s)

• P is oxidized as it haslost electrons (oxidation number 0 to +2);
• Ni is reduced as it has gained electrons (oxidation number 2+ to 0)

C2H4(g) + 3O2(g)⟶2CO2(g) + 2H2O(g)

• H2 is oxidized as it has lost electrons (oxidation number +2 to +1);
• C is reduced as it has gained electrons (oxidation number -2 to +4)

Zn(s) + H2SO4(aq) ⟶ZnSO4(aq) + H2(g)

• Zn is oxidized as it has lost electrons (oxidation number 0 to +2);
• H2 is reduced as it has gained electrons (oxidation number +2 to 0)

2K2S2O3(s) + I2(s)⟶ K2S4O6(s) +2KI(s)

• H2 is oxidized as it has lost electrons (oxidation number +2 to +1);
• C is reduced as it has gained electrons (oxidation number -2 to +4)

3Cu(s) + 8HNO3(aq) ⟶ 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

• Cu is oxidized as it has lost electrons (oxidation number 0 to +2);
• Nitrogen is reduced as it has gained electrons (oxidation number +5 to +4)