Electrolysis of Brine with a Limited Power Supply
A student sets up an electrolytic cell to produce chlorine gas from concentrated NaCl solution using inert electrodes. The power supply can deliver a maximum current of 2.00 A and a maximum voltage of 5.00 V. The student wants to produce 0.0100 mol of Cl2 gas in 30.0 minutes. The cell has a resistance of 3.50 Ω. The standard reduction potentials are: Cl2 + 2e^- → 2Cl^- E° = +1.36 V; 2H2O + 2e^- → H2 + 2OH^- E° = −0.83 V; Na+ + e^- → Na E° = −2.71 V. Overpotential for Cl2 evolution on the inert electrode is 0.20 V, and for H2 evolution is 0.10 V. The student must determine whether the power supply is adequate and what products will form.
1. Calculate the minimum voltage required for the electrolysis to produce Cl2 and H2, including overpotentials. Compare with the power supply's maximum voltage and determine if the cell can operate.
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At the anode, the possible oxidation reactions are: 2Cl^- → Cl2 + 2e^- (E°ox = −1.36 V) and 2H2O → O2 + 4H+ + 4e^- (E°ox = −1.23 V). With overpotential for Cl2 (0.20 V), the effective anode potential for Cl2 evolution is 1.36 + 0.20 = 1.56 V (vs SHE). For water oxidation, overpotential is typically higher, but not given; assuming no overpotential, it is 1.23 V. However, in concentrated NaCl, Cl2 evolution is favored due to high [Cl^-] and the overpotential for O2 is usually high. The problem states overpotential for Cl2 and H2 only. At the cathode, water reduction: 2H2O + 2e^- → H2 + 2OH^- E° = −0.83 V; with overpotential 0.10 V, the required cathode potential is −0.83 − 0.10 = −0.93 V. The minimum cell voltage (thermodynamic) = E_anode − E_cathode = 1.56 − (−0.93) = 2.49 V. Additionally, the cell resistance causes an ohmic drop: V_ohmic = I × R. At 2.00 A, V_ohmic = 2.00 × 3.50 = 7.00 V. Total voltage required = 2.49 + 7.00 = 9.49 V. This exceeds the 5.00 V maximum. Therefore, the power supply is not adequate to operate at 2.00 A. The student must reduce the current or resistance. At 5.00 V, the maximum current is (5.00 − 2.49)/3.50 = 0.717 A, assuming the thermodynamic voltage remains constant.
2. Calculate the maximum amount of Cl2 (in mol) that can be produced in 30.0 minutes with the power supply limited to 5.00 V and the cell resistance of 3.50 Ω. Use the current calculated from the voltage limit.
Read the worked answer
From the previous task, the maximum current at 5.00 V is I = (5.00 V − 2.49 V)/3.50 Ω = 0.717 A. The total charge passed in 30.0 min = 30.0 × 60 = 1800 s: Q = I × t = 0.717 A × 1800 s = 1290.6 C. The reaction for Cl2 production is 2Cl^- → Cl2 + 2e^-, so 2 mol e^- per mol Cl2. Moles of electrons = Q/F = 1290.6 C / 96485 C/mol = 0.01338 mol e^-. Moles of Cl2 = 0.01338 / 2 = 0.00669 mol. This is less than the target 0.0100 mol. Therefore, the student cannot produce the desired amount in 30.0 minutes with this power supply. To produce 0.0100 mol Cl2, required moles of e^- = 0.0200 mol, charge = 0.0200 × 96485 = 1929.7 C. At 0.717 A, time required = 1929.7 / 0.717 = 2691 s = 44.9 min. The student would need to extend the time or increase the voltage/current.