QUARTER FINAL STAGE 2026
PRESEC, Legon: 56 points
Mfantsipim School: 39 points
Presby SHS, Bompata: 24 points
QUESTION
A current of 9.65 amperes is passed through molten calcium chloride for 20.0 minutes.
Calculate the mass of calcium deposited.
ANSWER: $2.40\times10^{3}$ milligrams (i.e. 2.40 g)
SOLUTION:
$m = \dfrac{I \times t \times M}{z \times F}$
Where:
$m$ = mass of calcium deposited (g)
$I$ = current (9.65 A)
$t$ = time in seconds ($20.0 \times 60 = 1200\text{ s}$)
$M$ = molar mass of calcium ($40\text{ g/mol}$)
$z$ = valency of calcium ion ($2$ for $Ca^{2+}$)
$F$ = Faraday's constant ($96500\text{ C/mol}$)
$m = \dfrac{9.65 \times (20.0 \times 60) \times 40}{2 \times 96500}$
$m = \dfrac{9.65 \times 1200 \times 40}{2 \times 96500}$
$m = \dfrac{1200}{10000} \times \dfrac{40}{2}$
$m = 0.12 \times 20$
$m = 12\times10^{-2} \times 20$
$m = 2.40\text{ g}$
$m = 2.40\times10^{3}\text{ mg}$
NOTE: The working uses Ca = 40 and F = 96500 C/mol, which the question does not state.
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PRACTICE QUESTIONS
1. QUESTION: A current of 9.65 A is passed through copper(II) sulfate solution for 50.0 minutes. Calculate the mass of copper deposited. (Cu = 64)
ANSWER: 9.60 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 50.0\times60$
$t = 3000\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{9.65\times3000}{96500}$
$\dfrac{9.65}{96500} = \dfrac{1}{10000}$
$n(e^-) = \dfrac{3000}{10000}$
$n(e^-) = \dfrac{3}{10}$
$n(e^-) = 0.300\text{ mol}$
$n(Cu) = \dfrac{0.300}{2}$
$n(Cu) = \dfrac{3\times10^{-1}}{2}$
$n(Cu) = 0.150\text{ mol}$
$m = 0.150\times64$
$m = 15\times10^{-2} \times 64$
$m = 9.60\text{ g}$
2. QUESTION: A current of 19.3 A is passed through molten aluminium oxide (dissolved in cryolite) for 15.0 minutes. Calculate the mass of aluminium deposited. (Al = 27)
ANSWER: 1.62 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 15.0\times60$
$t = 900\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{19.3\times900}{96500}$
$\dfrac{19.3}{96500} = \dfrac{1}{5000}$
$n(e^-) = \dfrac{900}{5000}$
$n(e^-) = \dfrac{9}{50}$
$n(e^-) = 0.180\text{ mol}$
$n(Al) = \dfrac{0.180}{3}$
$n(Al) = \dfrac{18\times10^{-2}}{3}$
$n(Al) = 0.0600\text{ mol}$
$m = 0.0600\times27$
$m = 6\times10^{-2} \times 27$
$m = 1.62\text{ g}$
3. QUESTION: A current of 3.86 A is passed through molten magnesium chloride for 50.0 minutes. Calculate the mass of magnesium deposited. (Mg = 24)
ANSWER: 1.44 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 50.0\times60$
$t = 3000\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{3.86\times3000}{96500}$
$\dfrac{3.86}{96500} = \dfrac{1}{25000}$
$n(e^-) = \dfrac{3000}{25000}$
$n(e^-) = \dfrac{3}{25}$
$n(e^-) = 0.120\text{ mol}$
$n(Mg) = \dfrac{0.120}{2}$
$n(Mg) = \dfrac{12\times10^{-2}}{2}$
$n(Mg) = 0.0600\text{ mol}$
$m = 0.0600\times24$
$m = 6\times10^{-2} \times 24$
$m = 1.44\text{ g}$
4. QUESTION: A current of 9.65 A is passed through zinc sulfate solution for 20.0 minutes. Calculate the mass of zinc deposited. (Zn = 65)
ANSWER: 3.90 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 20.0\times60$
$t = 1200\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{9.65\times1200}{96500}$
$\dfrac{9.65}{96500} = \dfrac{1}{10000}$
$n(e^-) = \dfrac{1200}{10000}$
$n(e^-) = \dfrac{3}{25}$
$n(e^-) = 0.120\text{ mol}$
$n(Zn) = \dfrac{0.120}{2}$
$n(Zn) = \dfrac{12\times10^{-2}}{2}$
$n(Zn) = 0.0600\text{ mol}$
$m = 0.0600\times65$
$m = 6\times10^{-2} \times 65$
$m = 3.90\text{ g}$
5. QUESTION: A current of 19.3 A is passed through nickel(II) sulfate solution for 10.0 minutes. Calculate the mass of nickel deposited. (Ni = 59)
ANSWER: 3.54 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 10.0\times60$
$t = 600\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{19.3\times600}{96500}$
$\dfrac{19.3}{96500} = \dfrac{1}{5000}$
$n(e^-) = \dfrac{600}{5000}$
$n(e^-) = \dfrac{3}{25}$
$n(e^-) = 0.120\text{ mol}$
$n(Ni) = \dfrac{0.120}{2}$
$n(Ni) = \dfrac{12\times10^{-2}}{2}$
$n(Ni) = 0.0600\text{ mol}$
$m = 0.0600\times59$
$m = 6\times10^{-2} \times 59$
$m = 3.54\text{ g}$
6. QUESTION: A current of 3.86 A is passed through a chromium(III) salt solution for 75.0 minutes. Calculate the mass of chromium deposited. (Cr = 52)
ANSWER: 3.12 g
SOLUTION:
$m = \dfrac{I\times t\times M}{z\times F}$
$t = 75.0\times60$
$t = 4500\text{ s}$
$n(e^-) = \dfrac{I\times t}{F}$
$n(e^-) = \dfrac{3.86\times4500}{96500}$
$\dfrac{3.86}{96500} = \dfrac{1}{25000}$
$n(e^-) = \dfrac{4500}{25000}$
$n(e^-) = \dfrac{9}{50}$
$n(e^-) = 0.180\text{ mol}$
$n(Cr) = \dfrac{0.180}{3}$
$n(Cr) = \dfrac{18\times10^{-2}}{3}$
$n(Cr) = 0.0600\text{ mol}$
$m = 0.0600\times52$
$m = 6\times10^{-2} \times 52$
$m = 3.12\text{ g}$