Quantity of Electricity and the Faraday Constant – Electrolysis and Its Applications (Senior 3 Chemistry) | Ganzaa

Electrolysis and Its Applications

Lessons

Senior 3 · Chemistry

Electrolysis and Its Applications

5 lessons

Lesson 4 of 5

Quantity of Electricity and the Faraday Constant

25 min

How much metal for how much current?

A factory that silver-plates spoons must know how long to leave each batch in the tank: too short and the coat wears through, too long and expensive silver is wasted. In 1833 Michael Faraday found the rule that makes this possible: the mass of a substance produced at an electrode is proportional to the quantity of electricity passed. Since electricity is a flow of electrons, and every ion needs a fixed number of electrons, the calculation is a mole calculation.

Definition

Quantity of electricity (charge), $Q$ : the current multiplied by the time for which it flows.

$Q = I \times t$

$Q$ is in coulombs (C), $I$ in amperes (A) and $t$ in seconds (s). Example: 2.0 A for 50 s carries $2.0 \times 50 = 100\,\mathrm{C}$ .

Faraday constant, $F$ : the charge carried by one mole of electrons, $96\,500\,\mathrm{C\,mol^{-1}}$ . So $n(\mathrm{e^-}) = \frac{Q}{96\,500}$ .

Electrons per ion

The half-equation tells you how many moles of electrons give one mole of product.

Table 4.1: Electrons needed for one mole of product

Half-equation	Moles of e⁻ per mole of product	Charge per mole of product
$\mathrm{Ag^+ + e^- \rightarrow Ag}$	1	96 500 C
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$	2	193 000 C
$\mathrm{Al^{3+} + 3e^- \rightarrow Al}$	3	289 500 C
$\mathrm{2H^+ + 2e^- \rightarrow H_2}$	2	193 000 C
$\mathrm{2Cl^- \rightarrow Cl_2 + 2e^-}$	2	193 000 C
$\mathrm{4OH^- \rightarrow 2H_2O + O_2 + 4e^-}$	4	386 000 C

Worked example

Problem. A current of 2.00 A passes through copper(II) sulfate solution for 30.0 minutes. What mass of copper is deposited on the cathode? (Cu = 63.5; $F = 96\,500\,\mathrm{C\,mol^{-1}}$ )

Solution. $t = 30.0 \times 60 = 1800\,\mathrm{s}$ , so $Q = 2.00 \times 1800 = 3600\,\mathrm{C}$ .

$n(\mathrm{e^-}) = \frac{3600}{96\,500} = 0.03731\,\mathrm{mol}$

$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$ , so $n(\mathrm{Cu}) = \frac{0.03731}{2} = 0.01865\,\mathrm{mol}$ , and mass $= 0.01865 \times 63.5 = 1.18\,\mathrm{g}$ .

Worked example

Problem. (a) What volume of hydrogen, at room temperature and pressure, is given off when a current of 0.500 A passes through dilute sulfuric acid for 3860 s? (b) How long would 0.250 A take to deposit 2.16 g of silver from silver nitrate solution? (Ag = 108; molar volume $24\,\mathrm{dm^3\,mol^{-1}}$ )

Solution. (a) $Q = 0.500 \times 3860 = 1930\,\mathrm{C}$ ; $n(\mathrm{e^-}) = \frac{1930}{96\,500} = 0.0200\,\mathrm{mol}$ ; $n(\mathrm{H_2}) = 0.0100\,\mathrm{mol}$ ; $V = 0.0100 \times 24\,000 = 240\,\mathrm{cm^3}$ .

(b) $n(\mathrm{Ag}) = \frac{2.16}{108} = 0.0200\,\mathrm{mol}$ , which needs $0.0200\,\mathrm{mol}$ of electrons: $Q = 0.0200 \times 96\,500 = 1930\,\mathrm{C}$ .

$t = \frac{Q}{I} = \frac{1930}{0.250} = 7720\,\mathrm{s} \approx 129\,\text{minutes}$

Worked example

Problem. How much charge is needed to produce 1.00 kg of aluminium from molten aluminium oxide? (Al = 27)

Solution. $n(\mathrm{Al}) = \frac{1000}{27} = 37.0\,\mathrm{mol}$ . Each $\mathrm{Al^{3+}}$ needs 3 electrons: $n(\mathrm{e^-}) = 3 \times 37.0 = 111\,\mathrm{mol}$ .

$Q = 111 \times 96\,500 = 1.07 \times 10^{7}\,\mathrm{C}$

That huge charge is why aluminium smelters are built where electricity is cheap.

Warning

Common mistakes. (1) Time must be in seconds: convert minutes and hours first. (2) Do not forget the electrons in the half-equation: copper and hydrogen need 2 moles of electrons per mole, aluminium 3, oxygen 4. (3) The same charge passes through every cell in a series circuit, so the same number of moles of electrons reacts in each.

Tip

"It, over F, over electrons." $Q = I t$ ; divide by 96 500 for moles of electrons; divide by the electrons in the half-equation for moles of product.

✦

Practice

What charge flows when a current of 1.5 A passes for 2.0 minutes?
How many moles of electrons carry a charge of 9650 C?
What charge is carried by 0.20 mol of electrons?
What mass of silver is deposited by a current of 3.0 A flowing for 965 s? (Ag = 108)
A current of 0.50 A flows for 1930 s through zinc sulfate solution. What mass of zinc is deposited? (Zn = 65)
What volume of chlorine, at room temperature and pressure, is given off when 4825 C passes through brine? (Molar volume $24\,000\,\mathrm{cm^3\,mol^{-1}}$ )
How long, in seconds, must a current of 1.0 A flow to deposit 0.050 mol of copper?
Two cells in series, one with silver nitrate solution and one with copper(II) sulfate solution, are electrolysed together. 1.08 g of silver is deposited. What mass of copper is deposited in the other cell? Explain. (Ag = 108, Cu = 63.5)

Show answersHide answers

$Q = 1.5 \times 120 = 180\,\mathrm{C}$ .
$\frac{9650}{96\,500} = 0.100\,\mathrm{mol}$ .
$0.20 \times 96\,500 = 19\,300\,\mathrm{C}$ .
$Q = 2895\,\mathrm{C}$ ; $n(\mathrm{e^-}) = 0.0300\,\mathrm{mol} = n(\mathrm{Ag})$ ; mass $= 0.0300 \times 108 = 3.24\,\mathrm{g}$ .
$Q = 965\,\mathrm{C}$ ; $n(\mathrm{e^-}) = 0.0100\,\mathrm{mol}$ ; $n(\mathrm{Zn}) = 0.00500\,\mathrm{mol}$ ; mass $= 0.325\,\mathrm{g}$ .
$n(\mathrm{e^-}) = \frac{4825}{96\,500} = 0.0500\,\mathrm{mol}$ ; $n(\mathrm{Cl_2}) = 0.0250\,\mathrm{mol}$ ; $V = 600\,\mathrm{cm^3}$ .
$n(\mathrm{e^-}) = 0.10\,\mathrm{mol}$ ; $Q = 9650\,\mathrm{C}$ ; $t = \frac{9650}{1.0} = 9650\,\mathrm{s}$ .
$n(\mathrm{Ag}) = 0.0100\,\mathrm{mol}$ , so $0.0100\,\mathrm{mol}$ of electrons passed through both cells. $n(\mathrm{Cu}) = 0.00500\,\mathrm{mol}$ ; mass $= 0.00500 \times 63.5 = 0.318\,\mathrm{g}$ . Copper needs two electrons per atom, silver only one.

Sources and credits

Written by Ganzaa. The diagram, table, worked examples and Practice items were drawn and written by Ganzaa.

1How much metal for how much current?

2Electrons per ion

How much metal for how much current?

Electrons per ion