Question

1. A solution of NaCl is neutral, with an expected PH of 7. If electrolysis was carried out on 500 mL of an NaCl solution with a current of 0.500 A, how many seconds would it take for the PH of the solution to rise to a value of 9? (PH is related to H+ concentration by PH = -log [H+]).

Answer 1

Upon the electrolysis of the NaCl aqueous solution, the following processes take place:

anode: 2Cl^- $\rightarrow$ Cl₂ + 2e^-

cathode: 2H₂O + 2e^-$\rightarrow$ H₂ + 2OH^-.

As one can see, hydroxide anions OH^- are produced during the electrolysis. Therefore, the pH increases. At pH 9, the concentration of OH^- anions is:

$c(OH^-) = 10^{-(14-pH)} = 10^{-5}$ M.

This means that the number of the moles of OH^- produced during the electrolysis is:

$n(OH^-) = cV = 10^{-5} (M)·500·10^{-3}(L) = 5·10^{-6}$ mol.

The first Faraday's law states the relation between the number of the moles $n$ of the substance produced during the electrolysis and the charge $Q$ passed through the system:

$n = \frac{Q}{Fz}$ ,

where $F$ is the Faraday constant 96485 C/mol and $z$ is the number of the electrons participating in the reaction divided by the stoichiometric coefficient of the substance produced. The hydroxide anion has a stoichiometric coefficient 2, so $z=1$ . The charge passed through the solution is the product of the time and the current:

$Q = It$ .

Therefore, the time needed to raise the pH to 9 is:

$t = \frac{nF}{I} = \frac{5·10^{-6}(mol)·96485(C/mol)}{0.500 (A)} = 0.96$ s.

Answer: it would take 0.96 s for the pH of the solution to rise to a value of 9.