Electrolytic Cells and Electrolysis

An electrolytic cell consists of two electrodes dipped in an electrolyte (molten or aqueous) and wired to an external DC source. The source, not the chemistry, decides which electrode gains electrons and which loses them.

• Cathode — wired to the negative terminal, flooded with electrons, so reduction happens here: cations are reduced or deposited.
• Anode — wired to the positive terminal, electrons withdrawn, so oxidation happens here: anions are oxidised, or the electrode material itself may dissolve.

Compare this with a galvanic cell, where the spontaneous reaction itself sets the anode negative and the cathode positive. The polarities swap between the two cell types, but one thing never changes: oxidation is always at the anode, reduction is always at the cathode, in both. For molten NaCl, the cathode reaction is $\text{Na}^+ + e^- \rightarrow \text{Na}(l)$ and the anode reaction is $2\text{Cl}^- \rightarrow \text{Cl}_2(g) + 2e^-$. In aqueous solutions water itself can be preferentially oxidised or reduced instead of the dissolved ions, a qualitative nuance worth knowing even without full quantitative treatment.

Faraday's First Law: the mass deposited or liberated at an electrode is directly proportional to the charge passed, $w \propto Q$, so $w = ZQ$, where $Z$ is the electrochemical equivalent and $Q = It$ (current in amperes times time in seconds gives coulombs).

Faraday's Second Law: when the same charge passes through different electrolytes connected in series, the masses liberated are proportional to their chemical equivalent weights. Combining both laws with the Faraday constant, $F = 96{,}500\ \text{C mol}^{-1}$ (the charge on one mole of electrons), gives the working formula used for every numerical in this topic:

where: $w$ = mass deposited or liberated (g), $M$ = molar mass of the substance (g mol⁻¹), $n$ = number of electrons transferred per ion in the balanced electrode half-reaction, $I$ = current (A), $t$ = time (s), $F = 96{,}500\ \text{C mol}^{-1}$.

Problem: A current of 2 A is passed through molten aluminium oxide for 1 hour. Calculate the mass of aluminium deposited at the cathode. ($M_{Al} = 27\ \text{g mol}^{-1}$)

Step 1. Write the cathode half-reaction and read off $n$: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s)$, so $n = 3$.

Step 2. Convert time to seconds and find the charge: $t = 1\ \text{h} = 3600\ \text{s}$, so $Q = It = 2 \times 3600 = 7200\ \text{C}$.

Step 3. Apply the formula: $w = \dfrac{Q\,M}{nF} = \dfrac{7200 \times 27}{3 \times 96{,}500} = \dfrac{194{,}400}{289{,}500}$.

Step 4. Evaluate: $w \approx 0.67\ \text{g of aluminium}$ is deposited at the cathode.

Competitive stretch

The reverse question is just as common: how long must a current of 5 A flow through $\text{CuSO}_4$ solution to deposit 1.27 g of copper ($M_{Cu} = 63.5\ \text{g mol}^{-1}$)? Here $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}(s)$ gives $n = 2$, so rearrange to $t = \dfrac{w\,nF}{IM} = \dfrac{1.27 \times 2 \times 96{,}500}{5 \times 63.5} = \dfrac{245{,}110}{317.5} \approx 772\ \text{s}$, about 12.9 minutes.

• Swapping electrode polarity between galvanic and electrolytic cells. Oxidation is always at the anode and reduction always at the cathode in both — only the +/- sign of each electrode flips between the two cell types.
• Using the wrong $n$. Always write the balanced half-reaction first: $n = 1$ for Ag⁺, $n = 2$ for Cu²⁺, $n = 3$ for Al³⁺. Guessing $n$ from the formula alone is a common source of wrong answers.
• Forgetting to convert time to seconds before computing $Q = It$, since $F$ is defined on a per-coulomb-second basis; using minutes or hours directly throws the answer off by orders of magnitude.
• Assuming the cation with lowest reduction potential is always what deposits in aqueous electrolysis, without allowing that water itself may be oxidised or reduced preferentially — a qualitative point worth remembering even though full overpotential treatment is beyond this syllabus.

Ask the tutor: want to try a few more Faraday's-law numericals with different metals and gases, or work through how to predict which species gets discharged first in aqueous electrolysis?