Identify the metal and the products formed in each step, providing relevant chemical equations.

Step 1: Analyze the initial steps to identify the metal and key compounds.

• Step (i) describes roasting an ore, producing a pungent gas that turns acidified potassium dichromate (VI) green. This gas is sulfur dioxide ($\text{SO}_2$), indicating the ore is a sulfide.
• Step (ii) describes dissolving the residue in nitric acid to form crystals.
• Step (iii) describes heating these crystals, evolving a brown acidic gas (nitrogen dioxide, $\text{NO}_2$) and a colourless gas (oxygen, $\text{O}_2$), leaving a yellow solid. This is characteristic of lead(II) nitrate decomposition, where the yellow solid is lead(II) oxide ($\text{PbO}$).
• Step (v) describes heating the yellow solid with charcoal to form shiny beads. This is the reduction of lead(II) oxide to lead metal ($\text{Pb}$). Based on these observations, the metal is lead ($\text{Pb}$).

a)

• (i) The gas formed when the ore was roasted in air is sulfur dioxide, as it has a pungent smell and turns acidified potassium dichromate (VI) green. $$\boxed{\text{Sulfur dioxide}}$$
• (ii) When the crystals (lead(II) nitrate) were heated, a brown acidic gas and a colourless gas were evolved. These are nitrogen dioxide and oxygen, respectively. $$\boxed{\text{Nitrogen dioxide and Oxygen}}$$
• (iii) The yellow solid that remained after heating the crystals in step (iii) is lead(II) oxide. $$\boxed{\text{Lead(II) oxide}}$$
• (iv) The shiny beads formed in step (v) by heating lead(II) oxide with charcoal are lead metal. $$\boxed{\text{Lead}}$$

b) The yellow solid from procedure (iii) is lead(II) oxide ($\text{PbO}$). When melted and electrolysed using graphite electrodes:

• (i) At the cathode (negative electrode), lead ions are reduced to lead metal. At the anode (positive electrode), oxide ions are oxidized to oxygen gas.
  • At cathode: Shiny beads of lead metal are formed.
  • At anode: A colourless gas (oxygen) is evolved.
• (ii) The reaction at the anode involves the oxidation of oxide ions to oxygen gas. $$2\text{O}^{2-}(\text{l}) \to \text{O}_2(\text{g}) + 4\text{e}^-$$

c) The crystals formed in step (ii) are lead(II) nitrate ($\text{Pb(NO}_3)_2$). When dissolved in water and reacted with potassium iodide ($\text{KI}$) solution, a yellow precipitate of lead(II) iodide ($\text{PbI}_2$) is formed.
The ionic equation for this reaction is:

$$\text{Pb}^{2+}(\text{aq}) + 2\text{I}^-(\text{aq}) \to \text{PbI}_2(\text{s})$$

d) Assuming "solution from (f)" is a typo and refers to the lead(II) nitrate solution (from dissolving crystals from step (ii)). When sodium hydroxide solution is added drop by drop to lead(II) nitrate solution:

• A white precipitate is formed, which then dissolves in excess sodium hydroxide solution to form a colourless solution. (The white precipitate is lead(II) hydroxide, $\text{Pb(OH)}_2$, which is amphoteric and dissolves in excess strong base to form a soluble plumbate(II) complex.)

e) Assuming "solution from (f)" is a typo and refers to the lead(II) nitrate solution. When a piece of zinc foil is added to lead(II) nitrate solution:

• (i) Zinc is more reactive than lead, so it will displace lead from its salt solution. This is a displacement reaction (or redox reaction). $$\boxed{\text{Displacement reaction}}$$
• (ii) The ionic equation for the reaction is: $$\text{Zn}(\text{s}) + \text{Pb}^{2+}(\text{aq}) \to \text{Zn}^{2+}(\text{aq}) + \text{Pb}(\text{s})$$