Excess iron filings are added to aqueous tin(II) chloride, SnCl2. Iron displaces tin from the solution: Fe(s) + SnCl2(aq) → FeCl2(aq) + Sn(s) (a) Write the ionic equation for this reaction, omitting any spectator ions. [1] (b) Using your ionic equation, write separate half-equat…

Answer: ((a)) Fe(s) + Sn^2+(aq) → Fe^2+(aq) + Sn(s) ((b)) Oxidation: Fe → Fe^2+ + 2e^-. Reduction: Sn^2+ + 2e^- → Sn. ((c)) Sn^2+ ions are the oxidising agent, because they gain electrons (are themselves reduced), causing iron to lose electrons (be oxidised). ((d)) Iron: 0 to +2 (increases by 2). Tin: +2 to 0 (decreases by 2).. Part (a): Ionic equation Chloride ions, Cl^-, appear unchanged on both sides of the full equation (two on each side), so they are spectator ions and can be removed: Fe(s) + Sn^2+(aq) → Fe^2+(aq) + Sn(s) Part (b): Half-equations Splitting t…

IGCSE Chemistry 0620: Electron Transfer in the Displacement of Tin by Iron

Part (a): Ionic equation

Chloride ions, $\text{Cl}^-$ , appear unchanged on both sides of the full equation (two on each side), so they are spectator ions and can be removed:

$\text{Fe(s)} + \text{Sn}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Sn(s)}$

Part (b): Half-equations

Splitting the ionic equation into the electron-loss and electron-gain processes:

Oxidation (iron atoms lose electrons): $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$

Reduction (tin ions gain electrons): $\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$

Both half-equations involve the transfer of the same number of electrons ( $2e^-$ ), so they combine to give the overall ionic equation in part (a).

Part (c): Identifying the oxidising agent

An oxidising agent is the species that gains electrons (and is itself reduced), causing another species to lose electrons.

Here, $\text{Sn}^{2+}$ ions gain two electrons and are reduced to tin atoms. In doing so, they cause the iron atoms to lose electrons. So $\text{Sn}^{2+}$ ions are the oxidising agent.

(Iron is the reducing agent: it loses electrons and is itself oxidised, causing the $\text{Sn}^{2+}$ ions to be reduced.)

Part (d): Oxidation number changes

Iron starts as uncombined metal, oxidation number $0$ , and ends as $\text{Fe}^{2+}$ , oxidation number $+2$ :

$0 \longrightarrow +2 \quad (\text{increase of } 2)$

Tin starts as $\text{Sn}^{2+}$ , oxidation number $+2$ , and ends as uncombined metal, oxidation number $0$ :

$+2 \longrightarrow 0 \quad (\text{decrease of } 2)$

Iron’s oxidation number increasing confirms it is oxidised; tin’s oxidation number decreasing confirms it is reduced, consistent with the half-equations in part (b).

Final answers

(a) $\text{Fe(s)} + \text{Sn}^{2+}(aq) \rightarrow \text{Fe}^{2+}(aq) + \text{Sn(s)}$
(b) Oxidation: $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$ ; Reduction: $\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$
(c) $\text{Sn}^{2+}$ ions are the oxidising agent.
(d) Iron: $0$ to $+2$ (increase of 2); Tin: $+2$ to $0$ (decrease of 2).