Electrolytic Cells
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2 common types of qns:
- predict species discharged/ pdts formed at each electrode
- calculate amount of pdt formed
Predict species discharged/ pdts formed at each electrode
- Identify all redox active species present (ions, H2O and/ or active electrodes e.g. Cu, Fe)
- Cations attracted to cathode; anions attracted to anode
- Write E° for reactions involving each species
| E° | |
| Cathode | more + or less – species is discharged |
| Anode | more – or less + species is discharged |
| Example: In the electrolysis of a solution of potassium iodide with phenolphthalein indicator, a pink colour is observed near the cathode and a yellow-brown colour is observed near the anode. Deduce the species discharged and provide explanations for the observations.
Identify all redox active species present at each electrode. Cathode: K+ and H2O Anode: I– and H2O
Write E° for reactions involving each species Cathode: K+ (-2.92) and H2O (-0.83) Anode: I– (+0.54) and H2O (+1.23)
Species discharged at the: Cathode (less – ve) : 2H2O + 2e– → H2 + 2OH– OH– is produced at the cathode which gives a pink colour with phenolphthalein indicator. Anode (less +ve) : 2I– → I2 + 2e– I2 is produced at the anode which combines with I– in solution to form I3– which is yellow-brown. |
Calculate amount of pdt formed
| Q = I t = n F |
I: current (A); t: time (s); n: moles of electrons; F: Faraday’s constant (96500 Cmol–1)
- Determine moles of electrons transferred (n=It/F).
- From half-eqn, determine the moles of pdt using molar ratio
| Example: When a current of 1.0 A was passed through a solution of KI for a certain period of time, 240 cm3 of H2 was produced at the cathode under room conditions. a) Calculate the amount of I2 produced. b) Determine the duration for which the electrolysis was conducted. a) Overall reaction: 2H2O + 2I– → H2 + 2OH– + I2 Vm of gas at rtp = 24 dm3 = 24000 cm3 n(H2) = 240/24000 = 0.01 mol n(I2) = n(H2) = 0.01 mol
b) 2H2O + 2e– → 1H2 + 2OH– Each mole of H2 requires 2e– to form. n(e–) = 2 x 0.01 = 0.02 mol Q = I t = n F 1.0 x t = 0.02 x 96500 t = 1930 s = 32.2 min |
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