Mass deposited by electrolysis

Worked example

A current of 3.5 A passes for 92 min through a solution of Mᶻ⁺ (z = 2; M = 173.5 g mol⁻¹). What mass of metal is deposited?

1. 1. Charge from current and time: solve for Q

   $Q = It$

   = 19320 C

2. 2. Moles from charge (Faraday): solve for n

   $n = \dfrac{Q}{zF}$

   = 0.1001 mol

3. 3. Moles from mass: solve for m

   $m = nM$

   = 17.37 g

Answer: m = 17.4 g

Method

Q = It, n = Q/(zF), m = nM. Time must be in seconds for coulombs.

Common mistakes

• Each metal ion needs z electrons: n = Q/(zF).

Formulas used

• Charge from current and time

  $Q = It$

• Moles from charge (Faraday)

  $n = \dfrac{Q}{zF}$

• Moles from mass

  $n = \dfrac{m}{M}$

More electrochemistry questions

• ΔG° from E°
• Nernst equation
• E°cell from reduction potentials
• Equilibrium constant from E°cell
• Electrode potential away from standard conditions
• Latimer diagram: skipping an oxidation state
• Disproportionation constant from a Latimer diagram
• Daniell cell away from standard conditions