Question

for each reaction, write the chemical formulae of the oxidized reactants in the space provided. write the chemical formulae of the reduced reactants in the space provided.

2fe(s) + 3cucl₂(aq) → 2fecl₃(aq) + 3cu(s)
reactants oxidized:
reactants reduced:

pb(s) + hg(no₃)₂(aq) → hg(l) + pb(no₃)₂(aq)
reactants oxidized:
reactants reduced:

feso₄(aq) + zn(s) → fe(s) + znso₄(aq)
reactants oxidized:
reactants reduced:

Explanation:

Step1: Determine oxidation states

Oxidation state of \(Fe\) in \(Fe(s)\) is \(0\), in \(FeCl_{3}\) is \(+3\). Oxidation state of \(Cu\) in \(CuCl_{2}\) is \(+2\), in \(Cu(s)\) is \(0\). Oxidation state of \(Pb\) in \(Pb(s)\) is \(0\), in \(Pb(NO_{3})_{2}\) is \(+2\). Oxidation state of \(Hg\) in \(Hg(NO_{3})_{2}\) is \(+2\), in \(Hg(l)\) is \(0\). Oxidation state of \(Zn\) in \(Zn(s)\) is \(0\), in \(ZnSO_{4}\) is \(+2\). Oxidation state of \(Fe\) in \(FeSO_{4}\) is \(+2\), in \(Fe(s)\) is \(0\).

Step2: Identify oxidized and reduced reactants

Substance that loses electrons (oxidation state increases) is oxidized. Substance that gains electrons (oxidation state decreases) is reduced.

• For \(2Fe(s)+3CuCl_{2}(aq)\to2FeCl_{3}(aq)+3Cu(s)\): \(Fe\) goes from \(0\) to \(+3\) (oxidized), \(CuCl_{2}\) (\(Cu^{2 +}\) goes to \(0\)) is reduced.
• For \(Pb(s)+Hg(NO_{3})_{2}(aq)\to Hg(l)+Pb(NO_{3})_{2}(aq)\): \(Pb\) goes from \(0\) to \(+2\) (oxidized), \(Hg(NO_{3})_{2}\) (\(Hg^{2+}\) goes to \(0\)) is reduced.
• For \(FeSO_{4}(aq)+Zn(s)\to Fe(s)+ZnSO_{4}(aq)\): \(Zn\) goes from \(0\) to \(+2\) (oxidized), \(FeSO_{4}\) (\(Fe^{2+}\) goes to \(0\)) is reduced.

Answer:

• \(2Fe(s)+3CuCl_{2}(aq)\to2FeCl_{3}(aq)+3Cu(s)\): reactants oxidized: \(Fe\); reactants reduced: \(CuCl_{2}\)
• \(Pb(s)+Hg(NO_{3})_{2}(aq)\to Hg(l)+Pb(NO_{3})_{2}(aq)\): reactants oxidized: \(Pb\); reactants reduced: \(Hg(NO_{3})_{2}\)
• \(FeSO_{4}(aq)+Zn(s)\to Fe(s)+ZnSO_{4}(aq)\): reactants oxidized: \(Zn\); reactants reduced: \(FeSO_{4}\)