Background Electrochemistry

The relevant electrochemistry was developed in the Eh-pH diagram course notes and should be consulted.

Reminder on calculating E° or E: The potential difference or voltage generated by an electrochemical cell at a certain temperature is strictly a function of the composition of the cell, i.e. activities of the reactants and products. It does not depend on the charge that passes through that potential (i.e. nF). The energy associated with passage of the charge through the potential difference does depend on the amount that is passed: Energy = voltage x charge. But, the potential itself generated by the cell has nothing to do with the charge that is passed. Therefore DO NOT multiply E°’s by n numbers to calculate E° or E for a cell!

Electrowinning Equations (Faraday’s Law Relationships)

(1) Farady’s Law

Faraday’s law states that the number of moles of metal produced in an electrolysis is directly proportional to the charge passed. The constant of proportionality is nF, where n = moles of electrons per mole of metal produced (an integer) and F is the Faraday which is 96485 C/mole e-; a mole of electrons has a charge of 96485 C. (i.e. 6.02205 x 1023 e-/mol e- x 1.60218 x 10-19 C/e-.) Taking into account the fact that charge, q = current x time for fixed current (or the integral of I vs. t for a varying current) and that the moles of metal produced = mass/atomic weight leads to the formula provided.

Moles of metal plated  q (charge passed in the electrolysis) {1}

moles metal = nM = q/nF {2}

The units of q/nF are C/(mole e-/mol metal x C/mole e-) = mol metal. Charge passed at constant current is q = It. Moles of metal = M/AW where M is the mass of metal plated and AW is the atomic weight in g/mol. Then moles metal plated is,

nM = It/nF = M/AW {3}

Rearranging gives,

M = It AW {4}
nF

From Faraday’s law it is obvious that the lower n is, the less electricity that will be required per unit mass of metal plated. Some metal ions have more than one oxidation state. For typical copper electrowinning the cathodic half reaction is,

Cu+2 + 2e- = Cu (1)

Alternative leaching processes have been developed that form cuprous complexes:

Copper sulfides CuCl2-aq (2)

In this case the cathodic half reaction is,

CuCl2- + e- = Cu + 2Cl- (3)

which would use half the electricity. (No such process is currently commercially applied.)

(2) Current Efficiency

The simple formula above determines mass of metal plated for a given current and time, or vice versa. If the only cathodic (reduction) process operative is metal ion reduction to metal, then the formula gives an accurate indication of the mass of metal for a given current and time. However, other reduction half reactions may also occur simultaneously. These unwanted side reactions also consume electricity (current) and result in a lowered efficiency of use of current for metal plating. This leads to the idea of current efficiency. Current efficiency (CE) for metal plating then (or any electrolytic process), is the ratio of actual mass of metal plated to the theoretical mass based on Faraday’s law. It is usually given in %.

CE = actual mass metal plated x 100 {5}
theoretical mass expected

The theoretical mass is given by Faraday’s law.

CE = 100M = 100nFM {6}
It AW/nF It AW

Now M is actual mass of metal plated. For instance, calculate the current efficiency for the following conditions: 100 g of copper was plated in a copper electrolysis experiment using a constant current of 4 A for 23 hours. What was the current efficiency?

CE = 100 g Cu x 100 x 2 mol e-/mol Cu x 96485 C/mole e- = 91.7% {7}
(23 hr x 3600 sec/hr x 4 C/sec x 63.546 g Cu/mol Cu)

(3) Energy Efficiency

There is a theoretical minimum energy required to electroplate a metal. This is the thermodynamic minimum voltage times the charge passed at 100% current efficiency. Electrical work (energy) is voltage times charge (more generally, ∫Vdq if the voltage varies with charge passed, which equals Vq at constant voltage). By definition 1 VC = 1 J (1 volt·coulomb = 1 joule). In practice the actual voltage will be greater than the thermodynamic minimum for a number of reasons, and due to less than 100% current efficiency the charge passed may be greater than the theoretical minimum. On both counts the energy required will be greater than the theoretical limit.

Energy efficiency is the ratio of the theoretical energy required to the actual energy required, in percent. The theoretical voltage is E, i.e. the thermodynamic cell voltage. The theoretical charge required is given by Faraday’s law,

q = nMnF {8}

Hence the minimum energy requirement is,

We’rev = E nMnF units in J {9}

(Work and cell thermodynamic voltage are related by,

-G = nFE = we’rev {10}
units in J/mol; this is the difference between we’rev and We’rev.

where We’rev and we’rev are the electrical work under reversible conditions.* In electrowinning E is negative (G > 0; E < 0); the reaction as written is not favourable so we'rev E. The current is forced to go in the opposite direction to the natural tendency of the cell.)

CE = 100 nFM = 100 nFnM {11}
It AW q

* Reversible and irreversible processes are reviewed in the next section. For now suffice it to say that if an opposing voltage equal to E, where E < 0, (an electrolysis) is applied, then the thermodynamic tendency is just matched or just overcome and the reaction is exceedingly slow, i.e. reversible.

Since M/AW = nM and It = q,

q = 100nMnF {12}
CE

Then the actual energy input is,

-We’ = Eappl q = 100EapplnMnF {13}
CE

(-We' is a positive number.) The energy efficiency is:

EE = 100We’rev = 100E nMnF = E CE {14}
We’ 100Eappl nMnF/CE Eappl

Take an example again of 100 g of copper plated as above at a voltage of 2.0 V with 91.7% current efficiency. (If the cell is large and relatively little copper is plated, the applied voltage will be about constant, as will be the current.) E =
-0.89 V (= E° assuming standard conditions; 0.89 V = the necessary applied voltage to just overcome the thermodynamic negative cell voltage. (In reality the Nernst equation E would be required for a real cell with non-standard activities.)

EE = 0.89 x 91.7/ 2.0 = 40.8% {15}

This is not very high. Reasons for this will be explained later.

(4) Specific Energy Consumption

This is the actual energy requirement in units of energy per unit amount of metal plated (e.g. J/mol). The derivation above employed the energy consumption, i.e. Eappl q.

-We’ = 100Eappl nM nF {16}
CE

Divide both sides by the moles of metal (nM here) to get the specific energy consumption in J/mol (designated -we’):

-w’e = 100 nFEappl {17}
CE

A more conventional unit is kilowatt-hours per tonne of metal. A kWh is 1000 watts for 1 hour = 1000 J/sec x 3600 sec = 3.6 x 106 J. To convert the energy consumption number to kWh/t metal involves only unit conversions:

-w’e J x 1 kWh x 1 mol metal x 10-6 g {18}
mol 3.6 x 106 J AW g t

For copper this works out to -we’ J/mol x 0.0043713 kWh/t.
J/mol

For copper electrowon as above at 2 V and 91.7% current efficiency,

-w’e = 2 mol e- x 96485 C x 2.0 V = 4.209 x 105 J/mol Cu {19}
mol Cu mol e-
91.7 x 0.01

= 4.209 x 105 J/mol x 0.0043713 kWh mol / J t = 1840 kWh/t Cu {20}

In practice a typical energy requirement for copper EW is about 1900-2000 kWh/t.

(5) Metal Production Rate

Starting with Faraday’s law and current efficiency again,

q = 100nMnF {21}
CE

nM = q CE {22}
100nF

where nM is the moles of metal produced. Only CE% (e.g. 91.7%) of the total charge passed goes to plate metal. The charge at constant current is q = It.

nM = It CE {23}
100nF

Metal is plated onto both sides of a cathode starter sheet (e.g. a steel sheet in copper electrowinning). The total plating surface area for a number N cathodes is AcN, where Ac is the surface area per cathode sheet. Taking j as the current density in A/m2, the current being passed is j times the total plating area, i.e. I = jAcN.

nM = jAcNt CE {24}
100nF

i.e. j (C/sec m2) x surface area (m2) x time (sec) = charge (C). Next, rearrange to obtain:

nM = dnM = jAcN CE mol/sec {25}
t dt 100nF

AcN is the total plating surface area. This may be obtained in two ways, either using N to be the number of cathode starter sheets with Ac being the area of both sides combined, or with N being the number of plating surfaces and Ac being the surface area of just one side. Either way is equivalent. Regardless, the fact that metal is plated on two sides is factored in. The area of the narrow sides is negligible and little copper plates there since the electric field is rather diffuse at the sides anyway. In practice, edge strips may be used to prevent plating there. This makes removal of the plated metal sheets much easier.

To get plating rate in mass per unit time, multiply dnM/dt by appropriate conversion factors. This depends on the metal being plated since it involves the atomic weight. For tonnes per day:

dM = dnM mol x AW g x 10-6 t x 3600 sec x 24 h {26}
dt dt sec mol g h d

For copper,

dMCu = dnCu mol x 63.546 g x 10-6 t x 3600 s x 24 h = 5.49037 dnCu t Cu/day
dt dt s mol g h d dt
{27}

A typical EW cell would contain 60 cathodes 1 m wide x 1-1.2 m deep and 61 anodes. Copper is plated on both sides of the cathodes. Typical current densities range from 200-350 A/m2. For copper the cathode production rate for a cell with 60 cathodes per cell, each with length x width = 1 m x 1m, at 200 A/m2 current density and 91.7% CE is:

200 C x 2 m2 x 60 sheets x 0.917
dnM = sec m2 sheet {28}
dt 2 mol e- x 96485 C
mol mol e-

= 0.11405 mol/sec

dMCu/dt = 0.11405 x 5.49037 = 0.6262 t Cu/day

Note: the cathode sheet has dimensions of 1 m x 1 m. The plating area on one side is 1 m2. The plating area of the whole sheet is 2 m2; we plate on both sides. The calculation above allows estimation of copper production based on current and current efficiency. This aids in design of an actual EW tankhouse. The number of cells needed to achieve a desired production per year can be readily determined. How many cells of 60 cathodes each would be needed to achieve 50,000 t/yr of copper production under the conditions we have been using in the calculations above?

0.6262 t Cu/day/cell x 365 days/y x S cells = 50,000 t/y {29}

S = 218.8

We would need 219 cells. The total number of cathode sheets that must be employed then is 219 x 60 = 13,140. If a fully plated copper cathode is 0.5 cm thick on average and 1 m long on each side, and given that the density of copper is 8.92 g/cm3, the average weight of a cathode would be:

100 cm x 100 cm x 0.5 cm x 8.92 g/cm3 = 44.6 kg (0.0446 t) {30}

This means that 1.1211 x 106 sheets of copper have to be handled per year, or 3071 per day. If the current density was increased the number of cells required could be lower. Often cathode quality issues limit the current density.

How long would it take to plate a copper cathode to a thickness of 0.5 cm? The metal production rate equation can be rearranged to obtain time. Since the current is constant, dM/dt is constant as well, so the mass M plated over a specified time t equals dM/dt:

MCu = jAcN CE x 63.546 g/sec {31}
t 100 nF

t = 100 nF MCu sec {32}
jAcN CE x 63.546

where MCu in g. Now Ac is the area for a single sheet (on one side), i.e. 1 m2 in this case. N now is 1. (We are considering the time to grow a single cathode copper sheet.) For MCu = 44.6 kg = 44,600 g as above, plated at 200 A/m2 with 91.7% CE:

100 x 2mol e- x 96485 C x 44,600 g
t = mol mol e- = 7.385 x 105 sec {33}
200 C x 1 m2 x 91.7 x 63.546 g
sec m2 mol

= 8.55 days

Obviously the higher the current density, the shorter the plating time.

The five relationships above are summarized in the table below.

Table 1. Summary of the Faraday's Law relationships.