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.

mathrm{al}(s)+mathrm{febr}_{3}(a q)
ightarrow mathrm{albr}_{3}(a q)+mathrm{fe}(s)

mathrm{ca}(s)+mathrm{zncl}_{2}(a q)
ightarrow mathrm{zn}(s)+mathrm{cacl}_{2}(a q)

mathrm{feso}_{4}(a q)+mathrm{mg}(s)
ightarrow mathrm{fe}(s)+mathrm{mgso}_{4}(a q)

Explanation:

Step1: Determine oxidation states

Oxidation state of \(Al\) in \(Al(s)\) is \(0\), in \(AlBr_{3}\) is \(+ 3\). Oxidation state of \(Fe\) in \(FeBr_{3}\) is \(+3\), in \(Fe(s)\) is \(0\). Oxidation state of \(Ca\) in \(Ca(s)\) is \(0\), in \(CaCl_{2}\) is \(+2\). Oxidation state of \(Zn\) in \(ZnCl_{2}\) is \(+2\), in \(Zn(s)\) is \(0\). Oxidation state of \(Mg\) in \(Mg(s)\) is \(0\), in \(MgSO_{4}\) is \(+2\). Oxidation state of \(Fe\) in \(FeSO_{4}\) is \(+2\), in \(Fe(s)\) is \(0\).

Step2: Identify oxidized and reduced reactants

Substance that undergoes increase in oxidation state is oxidized. Substance that undergoes decrease in oxidation state is reduced.

For \(Al(s)+FeBr_{3}(aq)\to AlBr_{3}(aq)+Fe(s)\):

• \(Al\) goes from \(0\) to \(+3\) (oxidized).
• \(FeBr_{3}\) (since \(Fe\) goes from \(+3\) to \(0\)) is reduced.

For \(Ca(s)+ZnCl_{2}(aq)\to Zn(s)+CaCl_{2}(aq)\):

• \(Ca\) goes from \(0\) to \(+2\) (oxidized).
• \(ZnCl_{2}\) (since \(Zn\) goes from \(+2\) to \(0\)) is reduced.

For \(FeSO_{4}(aq)+Mg(s)\to Fe(s)+MgSO_{4}(aq)\):

• \(Mg\) goes from \(0\) to \(+2\) (oxidized).
• \(FeSO_{4}\) (since \(Fe\) goes from \(+2\) to \(0\)) is reduced.

Answer:

1. Reactants oxidized: \(Al\); Reactants reduced: \(FeBr_{3}\)
2. Reactants oxidized: \(Ca\); Reactants reduced: \(ZnCl_{2}\)
3. Reactants oxidized: \(Mg\); Reactants reduced: \(FeSO_{4}\)