Question

A current of $4.0\text{ A}$ is passed through $0.5\text{ L}$ of $0.2\text{ M NaCl}$ solution for $1200\text{s}$. Calculate the $\text{pH}$ of the solution after electrolysis.

• A. \( 1.3 \)
• B. \( 13 \)
• C. \( 7.0 \)
• D. \( 2.0 \)

Hint:

Whenever you notice that the electrolysis of a neutral salt solution releases \(\text{OH}^-\) or \(\text{H}^+\) ions, look immediately at the order of magnitude of the concentration to eliminate choices! Since \([\text{OH}^-] = 0.1\text{ M}\), the solution becomes distinctly basic, which means the \(\text{pH}\) must end up well above 7. This instantly rules out options A, C, and D in under two seconds, making Option B the only physically viable choice.

Answer 1

The Correct Option is B

Concept: During the electrolysis of an aqueous sodium chloride (\(\text{NaCl}\)) solution (brine), multiple species compete at the electrodes:
• At the Cathode (Reduction): Water has a higher reduction potential than sodium ions (\(\text{Na}^+\)). Therefore, water is preferentially reduced to produce hydrogen gas (\(\text{H}_2\)) and hydroxyl ions (\(\text{OH}^-\)): \[ 2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq) \]
• At the Anode (Oxidation): Chloride ions (\(\text{Cl}^-\)) are oxidized to form chlorine gas (\(\text{Cl}_2\)): \[ 2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^- \] As \(\text{H}_2\) and \(\text{Cl}_2\) escape, accumulation of \(\text{OH}^-\) ions together with spectator \(\text{Na}^+\) ions converts the remaining liquid environment into basic sodium hydroxide (\(\text{NaOH}\)), driving up the solution's overall \(\text{pH}\). According to Faraday's Laws, the moles of electrons passed dictate the exact moles of \(\text{OH}^-\) produced in a \(1:1\) stoichiometric ratio.

Step 1: Calculating the total charge and moles of electrons passed.
First, find the total electrical charge (\(Q\)) passed through the electrolytic cell using \(Q = I \cdot t\): \[ Q = 4.0\text{ A} \times 1200\text{ s} = 4800\text{ C} \] Now, compute the total number of moles of electrons passed by dividing by Faraday's constant (\(F \approx 96500\text{ C/mol}\)): \[ \text{Moles of } e^- = \frac{Q}{F} = \frac{4800}{96500} \approx 0.04974\text{ moles} \]

Step 2: Checking the limiting reactant condition.
Let's verify if there are enough initial \(\text{Cl}^-\) ions to sustain this electrolysis runtime: \[ \text{Initial moles of NaCl} = \text{Molarity} \times \text{Volume (L)} = 0.2\text{ M} \times 0.5\text{ L} = 0.1\text{ moles} \] Since the system requires only \(0.04974\text{ moles}\) of electrons to complete the reaction, the initial \(\text{NaCl}\) supply (\(0.1\text{ moles}\)) is in excess. The reaction is entirely controlled by the total charge passed. From the cathode half-reaction, \(1\text{ mole}\) of electrons yields exactly \(1\text{ mole}\) of \(\text{OH}^-\) ions: \[ \text{Moles of OH}^- \text{ produced} = \text{Moles of } e^- \approx 0.05\text{ moles} \]

Step 3: Calculating hydroxyl concentration, $\text{pOH}$, and final $\text{pH}$.
The molar concentration of hydroxyl ions produced in the \(0.5\text{ L}\) container volume is: \[ [\text{OH}^-] = \frac{\text{Moles of OH}^-}{\text{Volume (L)}} = \frac{0.05\text{ moles}}{0.5\text{ L}} = 0.1\text{ M} = 10^{-1}\text{ M} \] Now, computing the alkalinity index (\(\text{pOH}\)) using its logarithmic rule: \[ \text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(10^{-1}) = 1 \] Finally, calculating the system's ending chemical balance index (\(\text{pH}\)) at \(25^\circ\text{C}\): \[ \text{pH} = 14 - \text{pOH} = 14 - 1 = 13 \]