The corrosion of iron is an electrochemical process that involves the standard reduction potentials — Electrochemistry Chemistry Question
Problem Context
The corrosion of iron is an electrochemical process that involves the standard reduction potentials given here at 25 °C.
Fe2+(aq) + 2e– → Fe(s) Eo = –0.44 V
O2(g) + 4H+(aq) + 4e– → 2H2O(l) Eo = +1.23 V
Calculate the voltage for the standard cell based on the corrosion reaction. 2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l)
Model Answer
2Fe(s) → 2Fe2+(aq) + 4e– Eo = +0.44 V
O2(g) + 4H+(aq) + 4e– → 2H2O(l) Eo = +1.23 V
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) Eo = +1.67 V
Calculate the voltage if the reaction in Part a occurs at pH = 4.00 but all other concentrations are maintained as they were in the standard cell.
Model Answer
E = Eo – (RT / nF) ln (Fe2+[ ]2 / H+[ ]4 PO2)
= 1.67V – (8.314 J/mol⋅K)(298 K) / (96,500 J/V)(4 mol) ln (1 / (1.0 ×10–4)^4 (1))
= 1.67 – (0.00642) (+36.84) = 1.43 V
For the reaction Fe(OH)2(s) + 2e– → Fe(s) + 2OH–(aq), Eo = –0.88 V. Use this information with one of the given standard potentials to calculate the Ksp of Fe(OH)2.
Model Answer
Fe(OH)2(s) + 2e– → Fe(s) + 2OH–(aq) Eo = –0.88 V
Fe(s) → Fe2+(aq) + 2e– Eo = +0.44 V
Fe(OH)2(s) æ Fe2+(aq) + 2OH–(aq) Eo = –0.44 V
∆Go = –nFEo = –RT ln Ksp
ln Ksp = nFEo / RT
ln Ksp = (2 mol) (96,500J⋅V–1) (–0.44 V) / (8.314J⋅mol–1⋅K–1) (298 K) = –34.28
Ksp = 1.30 × 10–15
An iron object may be protected from corrosion by coating it with tin. This method works well as long as the tin coating is intact. However, when the coating is penetrated, the corrosion of the iron is actually accelerated. Use electrochemical principles to account for both of these observations. The standard reduction potential for tin is:
Sn2+(aq) + 2e– → Sn (s) Eo = –0.14 V
Model Answer
When iron is coated with Sn, the reaction Sn → Sn2+ + 2e– takes place. If the tin coating is broken, the reaction Sn2+ + Fe → Sn + Fe2+ becomes spontaneous. Iron becomes the anode and is oxidized more readily.