Question

What happens when : (a) An iron nail is dipped in copper (II) sulphate solution ?
(b) Potassium iodide solution is mixed with lead nitrate solution ?
(c) Silver chloride is exposed to sunlight ?
Write balanced chemical equations to support your answer.

Hint:

(a) Displacement: Fe + CuSO₄ \(\Rightarrow\) FeSO₄ + Cu (blue to green, red-brown coating) (b) Precipitation: Pb(NO₃)₂ + 2KI \(\Rightarrow\) PbI₂↓ (yellow) + 2KNO₃ (c) Photodecomposition: 2AgCl \(\Rightarrow\) 2Ag + Cl₂ (white to grey in sunlight)

Answer 1

What Happens When the Following Reactions Take Place:

(a) An Iron Nail is Dipped in Copper (II) Sulphate Solution:

Iron is more reactive than copper. Therefore, iron displaces copper from copper sulphate solution. This is an example of a displacement reaction.

Observations:
– The blue colour of copper sulphate solution gradually turns green due to the formation of iron (II) sulphate.
– A reddish-brown deposit of copper is formed on the iron nail.

Balanced Chemical Equation:
Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)

(b) Potassium Iodide Solution is Mixed with Lead Nitrate Solution:

When potassium iodide reacts with lead nitrate, a double displacement reaction occurs. A bright yellow precipitate of lead iodide is formed.

Observations:
– A yellow precipitate appears immediately.
– The precipitate is lead iodide (PbI₂).

Balanced Chemical Equation:
2KI (aq) + Pb(NO₃)₂ (aq) → PbI₂ (s) + 2KNO₃ (aq)

(c) Silver Chloride is Exposed to Sunlight:

Silver chloride undergoes photochemical decomposition when exposed to sunlight.

Observations:
– White silver chloride gradually turns grey.
– This happens due to the formation of metallic silver.

Balanced Chemical Equation:
2AgCl (s) → 2Ag (s) + Cl₂ (g)

Conclusion:
(a) Iron displaces copper from copper sulphate solution.
(b) Potassium iodide and lead nitrate form yellow lead iodide precipitate.
(c) Silver chloride decomposes in sunlight forming silver and chlorine gas.