Nonspontaneous processes are driven by application of an external power supply. The reactivity in an electrolytic cell is complicated by the fact that the solvent (usually water) and any dissolved species can react. The reaction that occurs is always the combination that leads to the least negative potential.
This means that to establish reactivity in an electrolytic cell, we must consider all possible combinations of oxidation/reductions reactions to determine the net cell reaction.
A 1 M solution of potassium iodide is electrolyzed under acidic conditions. What are the products?
KI(aq) → K+(aq) + I–(aq)
From the Table of Reduction Potentials, the possible reactions are:
K+(aq) + e– → K(s) E° = –2.92 V
I2(s) + 2 e– → 2 I–(aq) E° = +0.54 V
O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) E° = +1.23 V
2 H+(aq) + 2 e– → H2(g) E° = 0.00 V
The possible reductions are:
K+(aq) + e– → K(s)
2 H+(aq) + 2 e– → H2(g)
The reduction of hydrogen ion is less negative, so this will be the cathode reaction.
The possible oxidations are:
2 I–(aq) → I2(s) + 2 e–
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e–
The oxidation of iodide ion is less negative, so this will be the anode reaction.
The net reaction is
2 H+(aq) + 2 I–(aq) → I2(s) + H2(g)
The standard potential is
E° = 0.00 + (–0.54) = –0.54 V
Thus, at least 0.54 V will need to be applied to the cell in order to initiate reaction.
Electrolytic cells can be used to find the stoichiometry of reactions because the charge passed through the cell can measured easily.
The current (amps, A) in an electrical circuit is the rate of charge flow, i.e. coulombs per second. By measuring the current and the time of the electrolysis, the number of moles of electrons in a reaction can be measured.
nF = At
A = current in amps
t = time in seconds
F = Faraday's constant = 96485 C/mol
n = number of moles of electrons
Water is electrolyzed in a cell at 25 mA for 15 minutes. How many mL of oxygen gas are produced at 1.0 atm and 25 °C?
2 H2O(l) → O2(g) + 4 H+(aq) + 4 e–
Strategy: determine the number of moles of electrons consumed, use the stoichiometric relationship between moles of electrons and moles of oxygen, and then use the ideal gas law to find the number of mL of O2.
A = 25 mA × (1 A/1000 mA) = 0.025 A = 0.025 C/s
t = 15 min × (60 s/1 min) = 9.0×102 s
n = At/F = (0.025 C/s)(9.0×102 s)/(96485 C/mol) = 2.3×10–4 mol e–
mol O2 = ¼×mol e– = ¼×2.3×10–4 = 5.8×10–5 mol
V(O2) = n(O2)RT/P = (5.8×10–5 mol)(0.0821 L·atm/mol·K)(298 K)/(1.0 atm) = 0.0014 L = 1.4 mL O2