Question 18.PS.11: Electrolysis of Aqueous NaOH Predict the results of passing ......

The net cell reaction is 2 H_2O(\ell) → 2 H_2(g) + O_2(g). Hydrogen is produced at the cathode and oxygen is produced at the anode. The cell potential is – 1.23 V.

Strategy and Explanation   First, list all the species in the solution: Na^+, OH^-, and H_2O. Next, use Table 18.1 to decide which species can be oxidized and which can be reduced, and note the standard reduction potential of each possible half-reaction.

Reductions:

Na^+(aq) + e^- → Na(s)                 E^\circ_{cathode} = – 2.71  V

2 H_2O(\ell) + 2 e^- → H_2(g) + 2 OH^-(aq)                  E^\circ_{athode} = – 0.83  V

Oxidations:

4 OH^-(aq) → O_2(g) + 2 H_2O(\ell) + 4 e^-                 E^\circ_{anode} =+ 0.40  V

6 H_2O(\ell) → O_2(g) + 4 H_3O^+(aq) + 4 e^-                  E^\circ_{anode} = + 1.229  V

Water will be reduced to H_2 at the cathode because the standard reduction potential for this half-reaction is more positive. At the anode, OH^- will be oxidized because the standard reduction potential is smaller than that for water. The net cell reaction is

2 H_2O(\ell) → 2 H_2(g) + O_2(g)

and the cell potential under standard conditions is

E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode} = (- 0.83  V)  –  (+ 0.40  V) = – 1.23  V