Electrolysis Formula
Faraday's laws: Calculate mass deposited using current, time, and charge
Faraday's Laws of Electrolysis
Key Formulas
Q = I × t
Q = charge (coulombs), I = current (amperes), t = time (seconds)
moles e⁻ = Q / F
F = 96,485 C/mol (Faraday constant)
mass = (Q × MM) / (n × F)
n = electrons per ion, MM = molar mass
Step-by-Step Calculation
- Calculate total charge:
Q = I × t (convert time to seconds!)
- Find moles of electrons:
moles e⁻ = Q / 96,485
- Determine moles of substance:
Use stoichiometry from half-reaction
Example: Cu²⁺ + 2e⁻ → Cu, so moles Cu = (moles e⁻)/2
- Calculate mass:
mass = moles × molar mass
Common Half-Reactions
Metal Deposition (Cathode)
Cu²⁺ + 2e⁻ → Cu (n = 2)
Ag⁺ + e⁻ → Ag (n = 1)
Al³⁺ + 3e⁻ → Al (n = 3)
Ni²⁺ + 2e⁻ → Ni (n = 2)
Zn²⁺ + 2e⁻ → Zn (n = 2)
Gas Evolution
2H⁺ + 2e⁻ → H₂ (n = 2 per H₂)
2H₂O + 2e⁻ → H₂ + 2OH⁻
2Cl⁻ → Cl₂ + 2e⁻ (n = 2 per Cl₂)
4OH⁻ → O₂ + 2H₂O + 4e⁻
(n = electrons per formula unit)
Worked Examples
Example 1: Copper Electroplating
Problem: A current of 5.00 A passes through molten CuCl₂ for 2.00 hours. How much Cu is deposited?
Half-reaction: Cu²⁺ + 2e⁻ → Cu
Solution:
Step 1: Calculate charge
t = 2.00 h × 3600 s/h = 7200 s
Q = I × t = 5.00 × 7200 = 36,000 C
Step 2: Moles of electrons
moles e⁻ = 36,000 / 96,485 = 0.373 mol
Step 3: Moles of Cu
moles Cu = 0.373 / 2 = 0.187 mol
Step 4: Mass of Cu
mass = 0.187 × 63.55 g/mol
mass = 11.9 g Cu
Example 2: Silver Plating
Problem: How long will it take to deposit 10.0 g Ag using a 2.50 A current?
Half-reaction: Ag⁺ + e⁻ → Ag
Solution:
Step 1: Moles of Ag
moles Ag = 10.0 / 107.87 = 0.0927 mol
Step 2: Moles of electrons needed
moles e⁻ = 0.0927 × 1 = 0.0927 mol (n = 1 for Ag)
Step 3: Charge needed
Q = 0.0927 × 96,485 = 8947 C
Step 4: Time required
t = Q / I = 8947 / 2.50 = 3579 s
t = 3579 / 60 = 59.6 min
t = 59.6 min (about 1.0 hour)
Example 3: Hydrogen Gas Production
Problem: What volume of H₂ (at STP) is produced by electrolysis of water using 10.0 A for 30.0 minutes?
Half-reaction: 2H₂O + 2e⁻ → H₂ + 2OH⁻
Solution:
Q = 10.0 × (30.0 × 60) = 18,000 C
moles e⁻ = 18,000 / 96,485 = 0.187 mol
moles H₂ = 0.187 / 2 = 0.0933 mol
V = 0.0933 × 22.4 L/mol
V = 2.09 L H₂ at STP
Example 4: Current Efficiency
Problem: Electrolysis should deposit 5.00 g Cu, but only 4.50 g is obtained. Find current efficiency.
Solution:
Current efficiency = (actual mass / theoretical mass) × 100%
Efficiency = (4.50 / 5.00) × 100%
Efficiency = 90.0%
Some current may produce side reactions or H₂ evolution
Common Mistakes
Time in Wrong Units
Q = I × t requires time in SECONDS! Convert hours/minutes to seconds first.
Wrong Electron Count (n)
Cu²⁺ needs 2 electrons, Al³⁺ needs 3. Check the half-reaction!
Forgetting Faraday Constant
F = 96,485 C/mol (not 96,500!). Use accurate value for precision.
Multiple Products Possible
In aqueous solutions, water can be reduced (H₂) or oxidized (O₂) instead of intended products!