1.

1. What does Aluminum's position in the EMF series (E^o = -1.662V_SHE) tell us about reactivity when placed in the aqueous solution with respect to metals higher in the series?

It will be airodically dissolved (oxidized) more readily than metals higher in the series. All will spontaneously react (corrode) when combined with the ions of any of these metals above it, or when combined with any reactions above it providing the oxidized form of the reaction is present at the Al surface.

2. Still considering Al and having only the thermodynamic information available in the EMF series, what other half-cell reaction is expected to occur (assume the solution is deaerated acid containing no other oxidants)
   H⁺ + e^- -> 1/2 H₂. Explain why this other reaction is necessary in order for Al to corrode.

Corrosion cannot occur unless the electrons prdouced by the Al oxidation are consumed by another half-cell reaction, i.e., by a reaction that occur in the reduction direction. Otherwise, the number of electrons would increase in the system and this is a violation of charge neutrality.

3. Explain the source of the great corrosion resistance of Al when placed in an aqueous solution of neutral pH (hint: the answer is not to be found in the thermodynamic information available in the EMF series).

2.

1. With increasingly strong oxidants, E_corr is more active (positive) and then passive. Explain and illustrate the behavior on the polarization plot, being sure to label the axes and curves completely.



Oxidant B polarizes the metal to more (+) E in the active region where the metal oxidize at a higher rate but is not a strong enough oxidant to obtain a more (+) where the new corrosion potential is in the passive region. Oxidant C does this, i.e., it is a stronger oxidant.

2. Add to the plot in 2a the exchange current and the equilibrium potential of the metal dissolution reaction.
3. Choose on of the curves you drew in 2a as the reduction oxygen gas. Add to this curve its other branch, i.e., the oxidation of water to form oxygen gas.

3.

1. Is the cell
   Cu + 2H⁺ = Cu²⁺ + H₂
spontaneous as written, i.e., will Cu metal corrode in a tank of acid that contains only the water, protons, Cu²⁺ ions and inert anions? (No) Assume standard conditions and use the following information to calculate free energy change.

E^o = 0.34 V_SHE for Cu²⁺ + 2e^- = Cu and E^o = 0 V for H⁺ + e^- = 1/2 H₂. Show your work and state why the calculated free energy change supports your answer.

Cu2+ + 2e- = Cu 2 H+ + 2e- = H2 Cu + 2 H+ = Cu2+ + H2

0.34 VSHE 0.00 VSHE -0.34 V

A positive standard free energy change for the cell means it will not spontaneously occur from left to right (as written).



2. For the above Cu/Cu²⁺ half cell reaction, calculate its equilibrium potential for a 10^-3 M CuSO₄ solution. State any assumptions that you make in the calculation.

Assuming, a_Cu = [Cu²⁺] in mol/l, i.e., the activity coefficient,

4

1. The electrical energy providied by a spontaneous chemical reaction is partially dissapated by the motion of charge through the electrolyte to produce IR voltage drop. Use this concept to explain why corrosion rates are higher when anodes and cathodes are close together rather than far apart. Support your explanation by drawing a polarization plot showing two magnitudes of IR drop and their corresponding corrosion rates.

Since IR is a function of the length of the conductor, the IR voltage decreases as the corrosion circuit decreases in length; the latter occurs as the anodic and cathodic sites more closer together, the smaller IR means more electrical energy is available to drive the A & C reactions.

IR2>IR1 and I2corr<I1corr