What is the minimum voltage required for the electrolysis of 1.0 M NaCl in neutral solution? 2H₂O + 2Cl^- (1.0 M) → H₂(1 atm) + Cl₂(1 atm) + 2OH^-(1 × 10^-7 M)
Note that
2H₂O(l) + 2e^- → H₂(g) +2OH^-(aq) ,E° = -0.83 V,
2H₂O(l) + 2e^- → H₂(g) +2OH^-(aq) ,E = -0.41 V([OH^-] = 1 × 10^-7M) ,
O₂(g) + 4H⁺(aq) + 4e^- →2H₂O(l) ,E° = +1.23 V,
Cl₂(g) + 2e^- → 2Cl^-(aq) E° = +1.36 V,and
Na⁺(aq) + e^- → Na(s) E° = -2.71 V

Question

Quizplus · Accepted Answer

The voltage required for electrolysis can be calculated using the standard reduction potentials and the Nernst equation.

First, we identify the half-reactions occurring at the cathode and anode:

Cathode: 2H+ + 2e- --> H2 (E° = 0.00 V)
Anode: 2Cl- --> Cl2 + 2e- (E° = 1.36 V)

The overall cell potential is:

E°cell = E°cathode - E°anode
E°cell = 0.00 V - 1.36 V
E°cell = -1.36 V

Since we are in a neutral solution, we can assume that the concentrations of H+ and OH- are equal (10^-7 M). Therefore, we can use the Nernst equation to calculate the actual cell potential:

Ecell = E°cell - (RT/nF) * ln(Q)
Q = [H+]^2/[Cl-]^2
Ecell = -1.36 V - (0.0592 V/n) * log10(10^-14)
Ecell = -1.36 V + 0.0592 V
Ecell = -1.30 V

To drive the electrolysis, we need a voltage higher than -1.30 V. The minimum voltage required is the negative of this value:

Minimum voltage = 1.30 V

Therefore, the best choice is B) 1.77 V, which is higher than the minimum required voltage.

What Is the Minimum Voltage Required for the Electrolysis of 1.0