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

Part (a): Iron nail dipped in copper (II) sulphate solution
Step 1: Identify the type of reaction.
This is a displacement reaction (also called single displacement reaction). Iron is more reactive than copper according to the reactivity series of metals.
Step 2: Observation.

• The blue colour of copper sulphate solution gradually fades and becomes light green.
• A reddish-brown coating of copper metal is deposited on the iron nail.

Step 3: Explanation.
Iron displaces copper from copper sulphate solution because iron is more reactive than copper. Iron loses electrons to form \( \text{Fe}^{2+} \) ions (which are light green), while copper ions gain electrons to form copper metal.
Step 4: Balanced chemical equation.
\[ \boxed{\text{Fe (s) + CuSO}_4\text{ (aq) \(\Rightarrow\) FeSO}_4\text{ (aq) + Cu (s)}} \] Ionic equation: \[ \text{Fe (s) + Cu}^{2+}\text{ (aq) \(\Rightarrow\) Fe}^{2+}\text{ (aq) + Cu (s)} \]
Step 5: Final answer for part (a).
\[ \boxed{\text{The blue colour fades, and a reddish-brown coating of copper deposits on the nail.}} \] Part (b): Potassium iodide solution mixed with lead nitrate solution
Step 1: Identify the type of reaction.
This is a precipitation reaction (double displacement reaction) where two soluble salts react to form an insoluble salt (precipitate).
Step 2: Observation.

• A bright yellow precipitate is formed immediately.
• The precipitate is lead iodide (PbI₂).

Step 3: Explanation.
Potassium iodide (KI) and lead nitrate (Pb(NO₃)₂) exchange ions. Lead ions (Pb²⁺) combine with iodide ions (I⁻) to form insoluble lead iodide (PbI₂), which is yellow. Potassium ions (K⁺) and nitrate ions (NO₃⁻) remain in solution as potassium nitrate.
Step 4: Balanced chemical equation.
\[ \boxed{\text{Pb(NO}_3\text{)}_2\text{ (aq) + 2KI (aq) \(\Rightarrow\) PbI}_2\text{ (s) + 2KNO}_3\text{ (aq)}} \]
Step 5: Final answer for part (b).
\[ \boxed{\text{A bright yellow precipitate of lead iodide is formed.}} \] Part (c): Silver chloride exposed to sunlight
Step 1: Identify the type of reaction.
This is a photochemical decomposition reaction (photolysis). Light energy causes the decomposition of silver chloride.
Step 2: Observation.

• White silver chloride (AgCl) turns grey or darkens.
• The grey colour is due to the formation of silver metal.

Step 3: Explanation.
When exposed to sunlight, silver chloride decomposes into silver metal and chlorine gas. This reaction is used in black and white photography.
Step 4: Balanced chemical equation.
\[ \boxed{2\text{AgCl (s)} \xrightarrow{\text{sunlight}} 2\text{Ag (s)} + \text{Cl}_2\text{ (g)}} \]
Step 5: Final answer for part (c).
\[ \boxed{\text{White silver chloride turns grey due to decomposition into silver metal and chlorine gas.}} \]