Tutorial #11
Electrochemistry

e.g. What is the reduction potential for the hydrogen electrode in a solution having a pH of 4.00? The half reaction is 2 H⁺_(aq) + 2 e^- ® H_{2(g, 1 atm)}

If pH = 4.00, then [H⁺] = 10^-4 M. For this reaction, Q = p_H2 / [H⁺]² = 1 / 10^-8 = 10⁸
Thus, E = E^o - RT/nF ln(Q)
= 0.00 V - 8.314(298)/(2)(96500) ln(10⁸)
= -0.24 V

The lesson to learn here is that the Nernst equation applies to half reactions as well as to complete REDOX reactions!

Electrolytic Cells

These cells have E⁰_cell < 0, and so the reactions will not occur spontaneously. In other words, to make them occur, we have to apply an external force. We do this by applying a potential difference across the electrodes using an external power source.

e.g. for the reaction using the two half-reactions:

Fe⁺² + 2 e^- à Fe E⁰ = -0.440 V
2 Ag à 2 Ag⁺ + 2 e^- E⁰ = -0.800 V

overall: Fe⁺² + 2 Ag à Fe + 2 Ag⁺ E⁰_cell = -1.240 V

To make this reaction occur, we must apply a potential of at least -1.240 V. The potential must be applied so that electrons are being supplied to the electrode where reduction is occuring. This is always the cathode, and in this case it is the iron electrode.

Predicting the reactions at the electrodes

In an eletrolytic cell, there is usually more than one species that could be oxidized at the anode, and more than one species that could be reduced at the cathode. Not only are hte obvious species present (e.g. metals and metal ions), but there is also a solvent (usually water) and some other ions, many of which are oxidizable or reducible. You need a rule to predict what happens at each electrode - perhaps the metal ion you are trying to reduce will not get reduced!

The rule is: the reaction with the more positive half-reaction potential will occur preferentially

For example, suppose you were to electrolyze molten (melted) calcium chloride, CaCl_2(l). What is produced at each electrode?

Since the only two species present are Ca⁺² ions and Cl^- ions, the Ca⁺² must get reduced to Ca_(s) and the Cl^- must get oxidized o Cl_2(g):

Ca⁺² + 2 e^- à Ca_(s)         E^o= -2.87 V
2 Cl^- à Cl_2(g) +  2 e^-       E^o= -1.36  V
E^o_cell = -2.87 - 1.36 = - 4.23 V

This shows that (1) the reaction is not spontaneous, and you really would have to apply a potential to make it spontaneous and (2) the required potential is very high (i.e. very negative). This is primarily because of the difficulty of reducing an active metal like calcium.

For example, suppose you try to electrolyze an aqueous solution of calcium chloride, CaCl_2(aq). What is produced at each electrode?

Here, we have the additional species water. It is both reducible (to form hydrogen) and oxidizable (to form oxygen). At the anode, the possible reactions are:

(1) 2 Cl^- à Cl_2(g) +  2 e^-                               E^o= -1.36  V and
(2) 2 H₂O_(l) à O_2(g) + 4 H⁺_(aq) + 4 e^-       E^o= -1.23 V
The latter has the more positive potential, so we predict that the product at the anode is oxygen gas.

At the cathode, the possible reactions are:

(1) Ca⁺² + 2 e^- à Ca_(s)                                 E^o= -2.87 V and
(2) 2 H₂O_(l) + 2 e^- à H_2(g) + 2 OH^-_(aq)    E^o= -0.83 V
And we see that the reduction of water is favored, so hydrogen gas is produced at the cathode. The calcium and chloride are not affected!!

Corrosion
Essentially the oxidation of the metal, along with the reduction of water. Deposition of corrosion products such as Fe₂O₃ occurs via another REDOX process.

e.g. Fe à Fe⁺² + 2 e^- E⁰ = +0.440 V

i.e. the reaction tends to occur, and the resultant electrons have a potential of +0.440 V (i.e. +0.440 J/C).

To prevent corrosion, we push higher potential electrons back into the iron. This can be done passively, whereby a piece of a more active metal is electrically connected to the iron.

e.g. Mg à Mg⁺² + 2 e^- E⁰ = +2.356 V

in other words, the electrons from Mg oxidation are at a higher potential than those from Fe oxidation. The higher potential electrons thus "pushes" the lower potential electrons back into the iron, preventing its oxidation.

Electroplating

e.g. How much Zn_(s) (in g) can be electroplated onto a metal object from ZnSO_4(aq) using a current of 3.00 A for 1.5 h?

Zn⁺² + 2 e^- à Zn_(s)

here, i = 3.00 A = 3.00 C/s
t = 1.5 h = 5,400 s

thus, the total charge used = q = it = 3 C/s x 5,400 s = 16,200 C

but, q = nF, where n=moles electrons transferred

Thus, n = 16,200 C / (96487 C/mol e^-) = 0.1679 mol e^-

thus, moles Zn⁺² reduced = 0.1679 / 2 = 0.0839 mol Zn

thus, g Zn = 0.0839 mol x 65.39 g/mol = 5.49 g Zn