Quantitative calculations in electrolysis rely on Faraday's laws and fundamental relationships connecting mass, charge, current, molar mass, and stoichiometry.

1. Charge Equation: $$ Q = It $$ 2. Unified Mass Formula: $$ m = \frac{MIt}{vF} $$ Where;
m = Mass deposited or dissolved (g)
M = Molar mass of element (g/mol)
I = Current (A)
t = Time (s)
v = Valency / Number of electrons transferred per ion
F = Faraday's constant (96,500 C/mol)

3. Gas Volume at S.T.P. :
1 mole of any gas at S.T.P. occupies 22.4 dm³ (22,400 cm³)

1. Calculating Mass Deposited at the Cathode

Example 1: Calculate the mass of copper deposited at the cathode when a current of 2.5 A is passed through an aqueous solution of $\text{CuSO}_4$ for 40 minutes.
(Atomic mass of $\text{Cu} = 63.5 \text{ g/mol}$, $1 \text{ F} = 96,500 \text{ C/mol}$).

Write out parameters $$ I = 2.5 A $$ $$ t = \text{40 minutes}= 40 × 60 = 2,400 s $$ $$ M = 63.5 \text{g/mol} $$ $$ F = 96,500 \text{C/mol} $$ electrode half-reaction:
Cu²⁺_(aq) + 2e⁻ → Cu_(s)
Number of electrons transferred (v) = 2
$$ m = \frac{MIt}{vF} $$ $$m = \frac{63.5 × 2.5 × 2400} {2 × 96500} $$ $$ m = \frac{381000}{193000} $$ $$m ≈ 1.97 g $$ The mass of copper deposited is 1.97 g.

2. Calculating Time Required for Electrolysis

Example 2: How long (in hours) will it take to deposit 5.4 g of silver at the cathode from an aqueous silver nitrate ($\text{AgNO}_3$) solution using a steady current of 1.5 A?
($\text{Ag} = 108 \text{ g/mol}$, $1 \text{ F} = 96,500 \text{ C/mol}$).

Write out parameters $$ m = 5.4 g $$ $$ M = 108 g/mol $$ $$ I = 1.5 A $$ $$ F = 96,500 C/mol $$
electrode half-reaction
Ag⁺_(aq) + e⁻ → Ag_(s)
Number of electrons transferred (v) = 1
$$ m = \frac{MIt}{vF} $$ $$ t = \frac{mvF}{MI} $$ $$ t = \frac{5.4 × 1 × 96,500}{108 × 1.5} $$ $$ t = \frac{521,100}{162} $$ $$ t = 3,216.67 s $$ Convert seconds to hours $$ t = \frac{3,216.67}{3,600} $$ $$ t ≈ \text{0.89 hours (or 53.6 minutes)} $$ It will take approximately 0.89 hours(53.6 minutes).

3. Calculating Volume of Gas Liberated at S.T.P.

Example 3: Calculate the volume of oxygen gas liberated at the anode at S.T.P. when a current of 5.0 A is passed through dilute sulphuric acid for 32 minutes and 10 seconds. (Molar volume of gas at S.T.P. $= 22.4 \text{ dm}^3/\text{mol}$, $1 \text{ F} = 96,500 \text{ C/mol}$).

Write out parameters $$ I = 5.0 A $$ $$ t = (32 × 60) + 10 = 1,930 s $$ $$ F = 96,500 C/mol $$ Calculate total charge passed (Q): $$ Q = I × t = 5.0 × 1,930 = 9,650 C $$ Anode half-reaction for oxygen evolution
4OH⁻_(aq) → 2H₂O_(l) + O₂_(g) + 4e⁻

4 moles of electrons (4 F) produce 1 mole of O₂ gas.

Calculate moles of electrons passed $$ \text{Moles of e⁻ } = \frac{Q}{F} $$ $$ = \frac{9,650}{96,500} = \text{0.1 mol} $$ Calculate moles of O₂ produced $$ \text{Moles of O₂ }= \frac{0.1}{4} = 0.025 mol $$ Calculate volume at S.T.P. $$ \text{ Volume = Moles × Molar Volume} $$ $$ \text{Volume = 0.025 mol × 22.4 dm³/mol} $$ $$ \text{Volume = 0.56 dm³ (or 560 cm³)} $$ The volume of oxygen gas liberated at S.T.P. is 0.56 dm³.

4. Series Electrolysis (Faraday's Second Law)

Example 4: Two electrolytic cells containing aqueous $\text{CuSO}_4$ and $\text{AgNO}_3$ solutions respectively are connected in series. If 3.175 g of copper is deposited in the first cell, calculate the mass of silver deposited in the second cell. ($\text{Cu} = 63.5 \text{ g/mol}$, $\text{Ag} = 108 \text{ g/mol}$).

Determine Chemical Equivalents (E) $$ E = \frac{\text{Molar Mass}}{Valency} $$ For Cu (Cu²⁺): $$ E_{Cu} = \frac{63.5}{2} = 31.75 g/eq $$ For Ag (Ag⁺): $$ E_{Ag} = \frac{108}{1} = 108 g/eq $$ Apply Faraday's Second Law formula $$ \frac{m_{Ag}}{m_{Cu}} = \frac{E_{Ag}}{E_{Cu}} $$ $$ m_{Ag} = \frac{m_{Cu} × E_{Ag}}{E_{Cu}} $$ $$ m_{Ag} = \frac{3.175 × 108}{31.75} $$ $$ m_{Ag} = \frac{342.9}{31.75} $$ $$ m_{Ag} = 10.8 g $$ The mass of silver deposited in the second cell is 10.8 g.

5. Calculating Mass Deposited from Faradays

Example 5: 0.1 Faraday of electricity was passed through a solution of $\text{CuSO}_4$. What is the maximum weight of copper deposited on the cathode? $(\text{Cu} = 64 \text{ g/mol})$.

Electrode half-reaction: $$ Cu²⁺ + 2e⁻ → Cu $$ $$ \text{2 F deposits 64 g of Cu} $$ $$ \text{0.1F deposits xg of Cu} $$ $$ \text{Mass of Cu deposited}= \frac{0.1 ×64}{2} $$ $$ \text{Mass }= \text{3.2 g} $$ The weight of copper deposited is 3.2 g.

6. Series Calculation (Equal Quantity of Electricity)

Example 6: A quantity of electricity liberates 3.6 g of silver from its salt. What mass of aluminium will be liberated from its salt by the same quantity of electricity? $(\text{Ag}$ = 108 $\text{g/mol}$, $\text{Al}$ = 27 $\text{g/mol})$.

Calculate quantity of electricity required to liberate 3.6g of silver (Ag) $$ Ag⁺ + e⁻ → Ag $$ $$ \text{1 F = 108g} $$ $$ \text{xF = 3.6g} $$ $$ \text{Q}= \frac{3.6}{108} $$ $$ = 0.0333 \text{ F} $$ Since the same quantity of electricity is passed through $$ Al³⁺ + 3e⁻ → Al $$ $$\text{3 F liberates 27g of Al}$$ $$\text{0.0333 F liberates xg of Al}$$ $$ \text{Mass of Al }= \frac{0.0333 × 27}{3} = 0.27g $$

7. Calculating Relative Atomic Mass (Molar Mass)

Example 7: 0.222 g of a divalent metal is deposited when a current of 0.45 A is passed through a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. $(1\text{ F} = 96,500\text{ C})$.

Write out parameters $$ m = 0.222 g$$ $$ I = 0.45 A $$ $$ \text{t = 25 minutes = 25 × 60 = 1,500 s }$$ $$ \text{v = 2 (divalent) }$$ Total charge (Q) $$ Q = I × t = 0.45 A × 1,500 s = 675 C $$ $$ M = \frac{m × v × F}{Q} $$ $$ M = \frac{0.222 × 2 × 96,500}{675} $$ $$ M = \frac{42,846}{675} $$ $$ \text{M = 63.5 g/mol} $$ The relative atomic mass of the divalent metal is 63.5 g/mol (Copper).

8. Calculating Mass Deposited from Current and Time

Example 8: Calculate the mass of silver deposited when a current of 2.6 A is passed through a solution of a silver salt for 70 minutes. $(\text{Ag}$ = 108 $\text{ g/mol}$, 1$\text{ F} = 96,500\text{ C})$.

Total quality of electricity (Q) $$ I = 2.6 A $$ $$ t = 70 minutes = 70 × 60 = 4,200 s $$ $$ Q = I × t = 2.6 × 4,200 = 10,920 C $$ $$ v = 1 $$ Electrode half-reaction $$ Ag⁺ + e⁻ → Ag $$ $$ \text{mass deposited, m} = \frac{MIt}{vF} $$ $$ m = \frac{10,920 × 108}{96,500} $$ $$ m = 12.22 g $$ The mass of silver deposited is 12.22 g.

9. Calculating Current Required for Electrolysis

Example 9: Calculate the current that must be passed into a solution of aluminium salt for 1 hour 30 minutes in order to deposit 1.5 g of aluminium. $(\text{Al}$ = 27 $\text{ g/mol}$, $1\text{ F} = 96,500\text{ C})$.

Write out the parameters $$ t = 1 hr 30 min $$ $$= (1 × 3600) + (30 × 60) $$ $$ = 3,600 + 1,800 = 5,400 s $$ $$ m = 1.5g $$ Write Electrode half equation $$ Al³⁺ + 3e⁻ → Al $$ $$ v = 3 $$ $$ I = \frac{mvF}{MT} $$ $$ I = \frac{1.5 × 3 × 96500}{27 × 5400} $$ $$ I = 2.98\text{ A} $$ The current required is 2.98 A.

10. Gas Volume Under Non-Standard Conditions

Example 10: Calculate the volume of oxygen evolved at 285 K and 0.92 $\times 10^5 \text{ N/m}^2$ when a current of 3.5 A is passed through acidified water for 2.5 minutes. (Molar volume of gas at S.T.P. = 22.4 $\text{ dm}^3$, standard pressure = 1.01 $\times 10^5 \text{ N/m}^2$, 1$\text{ F} = 96,500\text{ C})$.

Total charge passed (Q) $$ I = 3.5 A $$ $$ t = 2.5 × 60 = 150s $$ $$ Q = I × t = 3.5 × 150 = 525 C $$ Calculate volume at S.T.P. $$ 4OH⁻ → 2H₂O + O₂ + 4e⁻ $$ $$ \text{4 F liberates 22.4 dm³ / 22,400 cm³} $$ $$ \text{V₁ (at STP)}= \frac{525 × 22,400 cm³}{4 × 96,500} $$ $$ = 30.47 cm³ $$ Apply General Gas Law $$ \frac{P₁ × V₁}{T₁} = \frac{P₂ × V₂}{T₂} $$ Where: $$ P₁ = 1.01 × 10⁵ N/m²$$ $$ T₁ = 273 K $$ $$ V₁ = 30.47 cm³ $$ $$ P₂ = 0.92 × 10⁵ N/m²$$ $$T₂ = 285 K$$ $$ V₂ = ? $$ $$ V₂ = \frac{P₁ × V₁ × T₂}{P₂ × T₁} $$ $$ V₂ = \frac{1.01 × 10⁵ × 30.47 × 285}{0.92 × 10⁵ × 273} $$ $$ V₂ = 34.92 cm³ $$ The volume of oxygen at 285 K and 0.92$ \times 10^5 \text{ N/m}^2$ is 34.92 cm³.

11. Comparative Moles in Series Electrolytic Cells

Example 11: 1.2 F of electricity are passed through electrolytic cells containing $\text{Na}^+, \text{Cu}^{2+}$, and $\text{Al}^{3+}$ in series. How many moles of each metal would be formed at the cathode of each cell?

For Sodium (Na⁺) $$ Na⁺ + e⁻ → Na $$ $$\text{1 F produces 1 mole of Na} $$ $$ \text{Moles of Na = 1.2 F × 1} $$ $$ = \text{1.2 moles of Na} $$ For Copper (Cu²⁺) $$ Cu²⁺ + 2e⁻ → Cu $$ $$ \text{2 F produces 1 mole of Cu} $$ $$ \text{1.2 F produces x mole of Cu} $$ $$ \text{Moles of Cu}= \frac{1.2}{2} $$ $$ = \text{0.6 moles of Cu} $$ For Aluminium (Al³⁺) $$ Al³⁺ + 3e⁻ → Al $$ $$ \text{3 F produces 1 mole of Al} $$ $$ \text{1.2 F produces x mole of Al} $$ $$ \text{Moles of Al }= \frac{1.2 F}{3} $$ $$ \text{= 0.4 moles of Al} $$ Answer:The moles formed are 1.2 moles Na, 0.6 moles Cu, and 0.4 moles Al.

12. Calculating Charge / Oxidation State on an Ion

Example 12: During electrolysis, a metal of relative atomic mass 27.02 is deposited if 0.1761 g of the metal is deposited on the cathode when a current of 0.15 A flows for a period of 3.52 hours. Calculate the charge on the ion of the metal. $$(1\text{ F} = 96,500\text{ C})$$.

Write out the parameters $$ I = 0.15\text{ A} $$ $$ t = 3.52 hrs = 3.52 × 3600 $$ $$ = 12,672 s $$ $$ M = 27.02 \text{ g/mol} $$ $$ m = 0.1761 \text{ g} $$ $$v = {MIt}{mF} $$ $$ v = \frac{27.02 × 0.15 × 12672}{0.1761 × 96,500} $$ $$ v = \frac{51,359.616}{ 16,993.65} $$ $$ v = 3.02 ≈ +3 $$ The charge on the metal ion is +3 (i.e., $\text{M}^{3+})$.

13. Calculating Cost of Electricity in Electrolysis

Example 13: If the cost of electricity required to deposit 1$\text{ g}$ of magnesium from a molten magnesium salt is $\text{₦}5.00$, how much would it cost to deposit $10\text{ g}$ of aluminium? $(\text{Mg}$ = $24\text{ g/mol}$, $\text{Al}$ = $27\text{ g/mol})$.

Determine the Quantity of electricity for Magnesium $$ Mg²⁺ + 2e⁻ → Mg $$ $$ \text{2 F deposits 24 g of Mg} $$ $$ \text{Q}= \frac{1 × 2}{24} = 0.08333\text{ F} $$ Determine Quantity of electricity required for 10 g of Aluminium (Al) $$ Al³⁺ + 3e⁻ → Al $$ $$ \text{3 F deposits 27 g of Al} $$ $$ \text{Q} = \frac{10× 3}{27} $$ $$ = \frac{30}{27}= 1.1111 F $$ Determine total cost using direct proportion $$ \text{Since 0.08333 F costs ₦5.00} $$ $$ \text{Cost for 10 g of Al} = \frac{5.00 × 1.1111}{0.08333} $$ $$ = ₦66.67 $$ The cost to deposit 10 g of aluminium is ₦66.67.