A2 Chemistry 9701: Quantitative Electrolysis - Q = It and the Faraday Constant

A steel spoon is electroplated with silver. The spoon is made the cathode in a cell containing aqueous silver nitrate, with a pure silver rod as the anode, and a constant current of $0.850\ \text{A}$ is passed for $15.0$ minutes.

(a) Write the half-equation for the reaction occurring at the cathode. [1]

(b) Calculate the quantity of electric charge, $Q$ , passed during the electroplating process. [1]

(c) Calculate the amount, in mol, of electrons transferred, and hence the mass of silver deposited on the spoon. ( $F = 96500\ \text{C mol}^{-1}$ ; $A_r(\text{Ag}) = 108$ ) [3]

(d) The same total charge calculated in (b) is then passed through a second electrolytic cell, connected in series with the first, containing molten aluminium oxide dissolved in cryolite (as in the industrial extraction of aluminium). Calculate the mass of aluminium that would be deposited at the cathode of this second cell by the same quantity of charge. ( $A_r(\text{Al}) = 27$ ) [3]

Part (a): Cathode half-equation

At the cathode, $\text{Ag}^+$ ions from solution are reduced, depositing silver metal onto the spoon:

$\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$

Part (b): Charge passed

First convert the time to seconds: $15.0\ \text{min} = 15.0 \times 60 = 900\ \text{s}$ .

$Q = It = 0.850 \times 900 = 765\ \text{C}$

Part (c): Moles of electrons and mass of silver deposited

$n(e^-) = \frac{Q}{F} = \frac{765}{96500} = 7.9275\times10^{-3}\ \text{mol} \approx 7.93\times10^{-3}\ \text{mol}$

From part (a), the mole ratio of electrons to silver deposited is $1:1$ (one electron reduces one $\text{Ag}^+$ ion), so:

$n(\text{Ag}) = n(e^-) = 7.9275\times10^{-3}\ \text{mol}$

$\text{mass of Ag} = n(\text{Ag}) \times A_r(\text{Ag}) = 7.9275\times10^{-3} \times 108 = 0.856\ \text{g}$

Check (independent recomputation): working directly from the fraction, $\dfrac{765 \times 108}{96500} = \dfrac{82620}{96500} = 0.8562\ \text{g}$ , consistent with the stepwise value of $0.856\ \text{g}$ .

Part (d): Mass of aluminium deposited by the same charge

The same charge, $Q = 765\ \text{C}$ , gives the same amount of electrons as in part (c): $n(e^-) = 7.9275\times10^{-3}\ \text{mol}$ (charge is unchanged because the two cells are in series, so the same current flows through both for the same time).

Aluminium is deposited via $\text{Al}^{3+}(l) + 3e^- \rightarrow \text{Al}(l)$ , a $3+$ ion, so three moles of electrons are needed per mole of aluminium:

$n(\text{Al}) = \frac{n(e^-)}{3} = \frac{7.9275\times10^{-3}}{3} = 2.6425\times10^{-3}\ \text{mol}$

$\text{mass of Al} = n(\text{Al}) \times A_r(\text{Al}) = 2.6425\times10^{-3} \times 27 = 0.0713\ \text{g} \ (71.3\ \text{mg})$

Check (independent recomputation): working directly from the fraction, $\dfrac{765 \times 27}{96500 \times 3} = \dfrac{20655}{289500} = 0.07135\ \text{g}$ , consistent with the stepwise value of $0.0713\ \text{g}$ . Note that far less aluminium than silver is deposited by the same charge, because each $\text{Al}^{3+}$ ion needs three electrons instead of one, and aluminium’s molar mass is also smaller.

Final answers

(a) $\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s)$
(b) $Q = \boxed{765\ \text{C}}$
(c) $n(e^-) = 7.93\times10^{-3}\ \text{mol}$ ; mass of Ag $= \boxed{0.856\ \text{g}}$
(d) $n(\text{Al}) = 2.64\times10^{-3}\ \text{mol}$ ; mass of Al $= \boxed{0.0713\ \text{g}}$ ( $71.3\ \text{mg}$ )