Module 6

Lesson 2.2  Operation of an Electrolytic Cell



Key Concepts


We can predict the half-reactions and the net reaction for an electrolytic cell by using the same procedure as was used for voltaic cells.

  1. Use the redox table to identify the SOA and the SRA. Do not forget to consider water for aqueous electrolytes.

  2. Write equations for the half-reactions. Include the reduction potential if required.

  3. Balance the electrons and write the net cell reactions, including the cell potential.

    \( \mathrm { E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} } \)

  4. If asked, state the minimum electric potential required (voltage) to force the reaction to occur. The minimum voltage is the absolute value of E°cell

Example

Consider an electrolytic cell containing cobalt(II) chloride solution and lead electrodes.

\( \mathrm { Pb(s)~ |~Co^{2+}(aq),~ Cl^-(aq)~ |~Pb(s) } \)

\( \mathrm { anode~|~electrolyte~|~cathode} \)











The minimum voltage required to operate this cell can be determined by applying the formula,    E°cell = E°cathode - E°anode    
Remember to use the reduction potentials!

\( \mathrm { E^{\circ}_{cell} = -0.28 V - (-0.13V) } \)

\( \mathrm { E^{\circ}_{cell} = -0.15 V } \)

According to the redox table, the minimum voltage required to operate this cell is 0.15 V.

The Chloride Anomaly

There are always exceptions to the rule! During the electrolysis of solutions containing the chloride ion, something unexpected occurs! Although water is the strongest reducing agent present, the chloride ions lose electrons at the anode. Remember that in cases where water and the chloride ion are the only reducing agents present, chlorine will lose its electrons preferentially (although it is a weaker reducing agent).

Fig. 3  Electrolysis of brine: reactions at the electrodes
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 Read pages 639 to 645 in the textbook.

Check Your Understanding

Complete Practice Questions 1 to 3 on page 640, Practice Questions 5 and 6 on page 644. Check your answers by clicking the link below.

Page 640 Practice Question 1

At the cathode of an electrolytic cell, the oxidizing agent gains electrons and is reduced. At the anode of the cell, the reducing agent loses electrons and is oxidized.


Page 640 Practice Question 2

Anode Cathode
Voltaic Cell - +
Electrolytic Cell + -


Page 640 Practice Question 3

In an electrolytic cell, electrons move through the external circuit from anode to cathode. Cations migrate toward the cathode and anions migrate toward the anode. This is the same as in a voltaic cell.


Page 644 Practice Question 5

  1. cathode: \( \mathrm { Ni^{2+}(aq) + 2e^- \rightarrow Ni(s) } \)
    anode: \( \mathrm { 2 I^-(aq) \rightarrow I_2(s) + 2e^- } \)
    net cell: \( \mathrm { Ni^{2+}(aq) + 2 I^-(aq) \rightarrow Ni(s) + I_2(s) } \)

    \( \mathrm { E^{\circ}_cell = -0.26 V - (+0.54 V) = -0.80 V } \)

    A minimum voltage of +0.80 V is required.

  2. cathode: \( \mathrm { (2 H_2O(l) + 2e^- \rightarrow H_2(g) + 2 OH^-(aq)) \times 2 } \)
    anode: \( \mathrm { 4 OH^-(aq) \rightarrow O_2(g) + 2 H_2O(l) + 4e^- } \)
    net cell: \( \mathrm { 2 H_2O(l) \rightarrow 2 H_2(g) + O_2(g) } \)

    \( \mathrm { E^{\circ}_{cell} = -0.83 V - (+0.40 V) = -1.23 V } \)

    A minimum voltage of +1.23 V is required.


Page 644 Practice Question 6

  1. \( \mathrm { E^{\circ}_{cell} = -0.41 V - (+1.07 V) = -1.48 V } \)

    The minimum voltage required is +1.48 V. Note that the SOA is Cr3+ and the SRA is Br-.

  2. \( \mathrm { E^{\circ}_{cell} = +0.34 V - (+0.34 V) = 0.00 V } \)

    The minimum voltage required is 0.00 V. Note that the SOA is Cu2+ and the SRA is Cu(s).