BACKGROUND RESEARCH

Update from my topic report on pH - currently in the EEI write-up

At standard pH [OH^-] = 1.0 x 10^-7 mol/L the reduction half reaction is written as:

O₂ + 2H₂O + 4e^- → 4OH^-

For writing the half reaction for acidic solutions H⁺ is the subject and the reaction must be corrected by adding 4H⁺ to each side of the equation, therefore cancelling out H₂O on both sides.

At pH [OH^-] = 1.0 x 10^-10 to 1.0 x 10^-8 mol/L the reduction half reaction is written as:

O₂ + 4H⁺ + 4e^- → 2H­₂O

At pH [OH^-] =1.0 x 10­^-11 or below the oxidation is now a result of hydrogen ions and not oxygen gas therefore the reduction half reaction is written as:

2H⁺ + 2e^- → H₂

The corrosion on Fe under all of these conditions should follow as:

2Fe_(s) + O_{2 (g)} + 2H₂O_(l) → 2Fe(OH)_2(s)

2Fe_(s) + O_{2 (g)} + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)

Fe_(s) + 2H⁺_(aq) → Fe²⁺_(aq)^­ + H­­_2(g)

This information demonstrates that as pH decrease (becomes more acidic) the half reaction for the corrosion on Fe changed from one that is initially driven by O ions to one that is driven by H ions. This is all within context as within tin can corrosion the Fe is under cathodic control undergoing the half reaction above while the Sn metal converts to Sn ions.

This information provides strong background theory into the corrosion of Sn within a tin can; however it all derives the following results and chemical half reactions as pH decreases as most organic matter in tin cans at a point is subject to anaerobic respiration where it forms an electrolyte of a low pH. If the pH was to alter and the reaction above take place within a basic electrolyte (pH <7) it would this affect the rate of corrosion and possibly the chemical half reactions taking place. At the same time it is not uncommon for electrode surface area (contrasting ratios on tin to iron) to vary within tin can corrosion and this poses yet another question that is viable for testing under these conditions.