Consider the half-reactions:
Cu^2+(aq) +2e^- ==> Cu(s) E naught = +0.34V
Sn^2+(aq) + 2e^- ==> Sn E naught = -0.14V
Fe^2+ + 2e^- ==> Fe(s) E naught = -0.44V
Al^3+(aq) = 3e^- ==> Al(s) E naught = -1.66V
Mg^2+(aq) = 2e^- ==> Mg(s) E naught = -2.37V
Which metals or metal ions will reduce Fe^2+(aq)?
The ion or metal with which you can obtain a + voltage when combined with the Fe^+2 + 2e ==> Fe(s)
Al and Mg
To determine which metals or metal ions can reduce Fe^2+(aq), we compare the reduction potentials (E°) of the half-reactions involved. In this case, we need to find metals or metal ions with reduction potentials more positive than -0.44V (the reduction potential for the Fe^2+ --> Fe(s) half-reaction).
Based on the given reduction potentials:
- Cu^2+(aq) + 2e^- ==> Cu(s) E° = +0.34V
- Sn^2+(aq) + 2e^- ==> Sn E° = -0.14V
- Fe^2+ + 2e^- ==> Fe(s) E° = -0.44V
- Al^3+(aq) + 3e^- ==> Al(s) E° = -1.66V
- Mg^2+(aq) + 2e^- ==> Mg(s) E° = -2.37V
We can see that the reduction potential of Cu^2+ is more positive than -0.44V. Therefore, Cu^2+ can reduce Fe^2+(aq). In other words, Cu metal can displace Fe from its aqueous ion form.
To summarize, copper (Cu) metal or copper ions (Cu^2+) can reduce Fe^2+(aq) to Fe(s).