Question 20.14: Calculating the Amount of Charge from the Amount of Product ...
Calculating the Amount of Charge from the Amount of Product in an Electrolysis
When an aqueous solution of copper(II) sulfate, \mathrm{CuSO}_{4}, is electrolyzed, copper metal is deposited.
\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)
(The other electrode reaction gives oxygen: 2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-}.) If a constant current was passed for 5.00 \mathrm{~h} and 404 \mathrm{mg} of copper metal was deposited, what was the current?
PROBLEM STRATEGY
The electrode equation for copper says that 1 \mathrm{~mol} \mathrm{Cu} is equivalent to 2 \mathrm{~mol} \mathrm{e}^{-}. You can use this in the conversion of grams of \mathrm{Cu} to moles of electrons needed to deposit this amount of copper. The Faraday constant (which says that one mole of electrons is equivalent to 9.65 \times 10^{4} \mathrm{C} ) converts moles of electrons to coulombs. The conversions are
\mathrm{g} \mathrm{Cu} \longrightarrow \mathrm{mol} \mathrm{Cu} \longrightarrow \mathrm{mol} \mathrm{e}^{-} \longrightarrow \text { coulombs }(\mathrm{C})
The current in amperes (A) equals the charge in coulombs divided by the time in seconds.
The conversion of grams of \mathrm{Cu} to coulombs required to deposit 404 \mathrm{mg} of copper is
0.404 \cancel{\mathrm{gCu}} \times \frac{1 \cancel{\mathrm{mol Cu}}}{63.6 \cancel{\mathrm{g Cu}}} \times \frac{2 \cancel{\text { mol e }}}{1 \cancel{\text { mol Cu }}} \times \frac{9.65 \times 10^{4} \mathrm{C}}{1 \cancel{\text { mol e }}}=1.2 \underline{2} 6 \times 10^{3} \mathrm{C}
The time lapse, 5.00 \mathrm{~h}, equals 1.80 \times 10^{4} \mathrm{~s}. Thus,
\text { Current }=\frac{\text { charge }}{\text { time }}=\frac{1.226 \times 10^{3} \mathrm{C}}{1.80 \times 10^{4} \mathrm{~s}}=\mathbf{6 . 8 1} \times \mathbf{1 0}^{-\mathbf{2}} \mathbf{A}