# Electrolysis

Skills to Develop

• Describe the chemical reactions in electrolysis.
• Calculate quantities used or generated in electrolysis.

Chemical reactions in batteries or galvanic cells provide the driving force for electrons to struggle through loads. This is how chemical energy is transformed into electric energy. Electrolysis can be carried out in solutions or molten salts (liquid). Because the atoms and ions have to move physically, the medium has to be a fluid. The products, like the reactants in a galvanic cell, can be in a solid, liquid, or gas state.

## Electrolysis of Molten Salts

Electrolysis is a process by which electrons are forced through a chemical cell, thus causing a chemical reaction. The positive charge usually attracts electrons, and the electrode providing electrons is called the cathode, because reduction takes place on it.

Reduction always takes place at the cathode, by definition. In the electrolysis of molten salt, $$\ce{NaCl}$$, the cathode and anode reactions are:

Cathode (reduction): $$\ce{Na+ + e^- \rightarrow Na}$$
Anode (oxidation): $$\ce{2 Cl- \rightarrow Cl2 + 2 e^-}$$

or

Anode oxidation $$\Delta E$$ Cathode reduction
$$\ce{2 Cl- \rightarrow Cl2 + 2 e^-}$$ e- pump $$\ce{2 Na+ + 2 e^- \rightarrow 2 Na}$$
$$\ce{2 Cl- + 2 Na+ \rightarrow Cl2 + 2 Na}$$

If one mole of electrons (96485 C or 1 Faraday) passes from the anode to the cathode, one mole of $$\ce{Na}$$ (23 g) will be deposited, and half a mole of chlorine gas $$\ce{Cl2}$$ (or one mole of $$\ce{Cl}$$ atoms) will be collected from the reaction at the anode. In the above setup, if the current is 1.0 A, the rate at which $$\ce{Na}$$ deposits at the cathode will be $$1.0 \times 10^{-5}$$ (= 1/96,485) mol (or 0.24 mg) per second. Of course, it takes 96,485 seconds (or 26.8 hrs) to deposit one mole (23 g) of sodium metal.

## Electrolysis of Water

Pure water does not conduct electricity well, because the numbers of $$\ce{H+}$$ and $$\ce{OH-}$$ ions are small (10-7 mol/L each). However, in the presence of an acid, water can be decomposed. A potential of -2.06 V is the standard cell potential for

$$\mathrm{Pt \,|\, H_2O,\: [H^+] = 1\: M \,|\, O_2 \,||\, H_2O\: [OH^-] = 1\: M \,|\, H_2 \,|\, Pt}$$

And when a potential greater than 2.06 V is applied such that the forward cell has a positive potential, the following reactions take place.

Anode oxidation $$\Delta E$$ Cathode reduction
$$\ce{H2O \rightarrow 4 H+ + 4 e^- + O2}$$ >2.06 V $$\ce{4 H2O + 4 e^- \rightarrow 2 H2 + 4 OH-}$$
$$\ce{2 H2O \rightarrow 2 H2 + O2}$$

Note that the acid and base must be separated by a salt bridge to prevent the neutralization reaction. This cell potential is different from the cell potential for the reverse reaction in pure water, for which $$\mathrm{[H^+] = [OH^-] = 10^{-7}}$$, and E° is 1.23 V.

## The Hall Process

Aluminum ($$\ce{Al}$$) is the third most abundant element on Earth's crust, in the form of bauxite or alumina $$\ce{Al2O3}$$. Because it is very reactive, this metal remained unknown to mankind until 1827. By then, Wohler obtained some $$\ce{Al}$$ metal by reducing $$\ce{Al2O3}$$ with potassium vapor. In 1886, two young men working two continents apart electrolyzed molten cryolite $$\ce{Na3AlF6}$$ (melting point 1000° C). Aluminum was not produced when pure cryolite was used. Electrolysis is successful only if the ions move to the electrodes, and the reactions take place. That was not the case for molten cryolite. Figure 1: Cryolite from Ivigtut Greenland. Image used with permission from Wikipedia

However, both Hall and Heroult tried to mix about 5% alumina in their molten cryolite and their discovery is now known as the Hall-Heroult process, which is a commercially viable process. In a modern process, the reactions are:

$$\ce{AlF6^3- + 3 e^- \rightarrow Al + 6 F-} \hspace{15px} \ce{Cathode}\\ \mathrm{\underline{2 Al_2OF_6^{2-} + C_{\large{(s)}} + 12 F^- \rightarrow + 4 AlF_6^{3-} + CO_2 + 4 e^- \hspace{15px} Anode}}\\ \mathrm{2 Al_2O_3 + 3 C \rightarrow 4 Al + 3 CO_2 \hspace{15px} Overall\: cell\: reaction}$$

The overall reaction is simple despite the complicated mechanism in the electrolysis. In Example 1 below, we use simple formulation for the cathode and anode reaction to illustrate the shoichiometry of electrolysis.

## Electrochemical Stoichiometry

Electrolysis causes chemical reactions. Amounts of reactants, products, energy, and charge are inter-related. The following examples illustrate the stoichiometry of electrolysis.

Example 1

The Hall process can be oversimplified by these reactions,

$$\mathrm{Al^{3+} + 3 e^- \rightarrow Al\hspace{20px}Cathode}\\ \mathrm{C_{\large{(s)}} + 2 O^{2-} \rightarrow CO_2 + 4 e^- \hspace{20px}Anode}$$

How many Faradays and how many coulombs must be passed through a molten mixture of $$\ce{Al2O3}$$ and $$\ce{Na3AlF6}$$ to produce 1 kg of $$\ce{Al}$$ metal?

SOLUTION

The reactions are unrealistic because the ions containing $$\ce{Al}$$ are not bare $$\ce{Al^3+}$$ ions. However, we use the simplified reaction for stoichiometry relationships only.

Study the following conversion method to get from 1 kg of $$\ce{Al}$$ to number of Faradays and coulombs. Note that values in the numerators are equivalent to those in the denominators in the factors.

\begin{align} 1\: &\mathrm{kg\: Al \:\dfrac{1000\: g}{1\: Kg} \:\dfrac{1\: mol}{26.98\: g} \:\dfrac{3\: F(araday)}{1\: mol\: Al}}\\ &= \ce{111\: M\: \dfrac{96485\: C}{1\: M}}\\ &= \mathrm{1.1 \times 10^{7}\: C} \end{align}

Producing $$\ce{Al}$$ is an expensive process.

Example 2

Thirty minutes (30 m) of electrolysis of a solution of $$\ce{CuSO4}$$ produced 3.175 g $$\ce{Cu}$$ at the cathode. How many Faradays and how many coulombs passed through the cell? What is the current?

SOLUTION

Using the same method as indicated above, you have

\begin{align} 3.175 &\mathrm{\:g\: Cu\: \dfrac{1\: mol}{63.5\: g} \:\dfrac{2\: F}{1\: mol\: Cu}\: (At.wt.\: Cu = 63.5)}\\ &= \mathrm{0.100\: M\: \dfrac{96485\: C}{1\: M}}\\ &= \mathrm{9650\: C} \end{align}

To calculate the current, you divide the charge (C) by the time period (sec).

\begin{align} \ce{I} &= \mathrm{\dfrac{9650\: C}{(30\times60\: sec.)}}\\ &= \mathrm{5.36\: A.} \end{align}

Example 3

An electrolysis cell with $$\ce{Fe(NO3)3}$$ solution is operated for 2.0 hrs at a constant current of 0.10 A; how much $$\ce{Fe}$$ metal is plated out if the efficiency is 90%? (At.wt. $$\mathrm{Fe=55.8}$$)

SOLUTION

The charge passed the cell is

$$\mathrm{0.10\: (C/sec)\times2\times3600\: sec = 720\: C.}$$

$$\mathrm{720\: C \:\dfrac{1\: M}{96485\: C} \:\dfrac{1\: mol\: Fe}{3\: M} \dfrac{55.8\: g\: Fe}{1\: mol\: Fe}\: 0.90}\\ \hspace{10px}= \mathrm{0.12\: g\: Fe}.$$

Note that the last factor corresponds to 90% efficiency.

Example 4

An electrolysis cell containing $$\ce{MSO4}$$ solution is operated for 1.0 hr at constant current of 0.200 A. If the current efficiency is 95%, and 0.399 g of $$\ce{M}$$ plates out, what is the atomic weight of the element $$\ce{M}$$?

SOLUTION

Taking the current efficiency of 95% into consideration, the effective charge passed through the cell is

$$\mathrm{0.200\: (C/sec) \times 3600\: sec \times 0.95 = 684\: C}$$

$$\mathrm{684\: C \:\dfrac{1\: M}{96485\: C} \:\dfrac{1\: mol\: M}{2\: M}} = \textrm{3.54 \times 10^{-3} mol M}$$

The atomic weight of $$\ce{M}$$ is thus,

$$\mathrm{\dfrac{0.399\: g}{\textrm{3.54\times 10^{-3}}} = 112.7\: g/mol}$$

Checking the results against a table indicates that the element is cadmium $$\ce{Cd}$$.

## Electroplating

Electroplating differs from electrolysis in that the metal deposited from electrolysis plates out on the surface of another metal. The electrolyte contains the plating metal in the form of dissolved ions and the anode usually is made of the plating metal. The object to be plated is the cathode. Electroplating technology is frequently use in metal finishing, metallic coating and finishing, and salt water pool chlorination.

## Questions

1. In an electrolysis, at which electrode does oxidation take place, anode or cathode?
2. In the electrolysis of water, at which electrode will hydrogen be produced, anode or cathode?
3. If a current of 1.00 A is used in the electrolysis of $$\ce{H2O}$$, how many seconds will it take to produce 22.4 mL $$\ce{H2}$$ at STP?

## Solutions

1. Answer . . . anode
Hint...
Convention in cell notation: oxidation anode cell is on the left, reduction cathode cell is on the right.

2. Answer . . . cathode
Hint...
Just another question to help you to remember that the cathode provides electrons for the reduction reaction,
$$\ce{2 H+ + 2 e^- \rightarrow H2}$$.

3. Answer . . . 96
Hint...
Facts: Two Faradays are required to produce 22.4 L $$\ce{H2}$$ at STP. Two mili-Faradays (0.002 mole) are required to produce 22.4 mL, 1 mili-mole or 0.001 mole, $$\ce{H2}$$ at STP.

$$\mathrm{22.4\: mL \:\dfrac{1\: mol}{22.4\: L} \:\dfrac{2\times96485\: C}{1\: mol} \:\dfrac{1\: s}{1\: C} = 193\: s}$$

Note: $$\mathrm{1\: A = 1\: C/s}$$ ; $$\mathrm{\dfrac{1\: s}{1\: C} = \dfrac{1}{(1\: A)}}$$