# 19.10: Counting Electrons


## Quantitative Considerations

If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.

The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction Ag+(aq) + e → Ag(s), 1 mol of electrons reduces 1 mol of Ag+ to Ag metal. In contrast, in the reaction Cu2+(aq) + 2e → Cu(s), 1 mol of electrons reduces only 0.5 mol of Cu2+ to Cu metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,486 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (C) transferred is the product of the current (A) and the time (t, in seconds):

$C = A \times t \label{20.9.14}$

The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.

For example, if a current of 0.60 A passes through an aqueous solution of CuSO4 for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:

$$\textrm{charge}=\textrm{(0.60 A)(6.0 min)(60 s/min)}=\mathrm{220\;A\cdot s}=\textrm{220 C} \label{20.9.15}$$

The number of moles of electrons transferred to $$Cu^{2+}$$ is therefore

$$\textrm{moles e}^-=\dfrac{\textrm{220 C}}{\textrm{96,486 C/mol}}=2.3\times10^{-3}\textrm{ mol e}^- \label{20.9.16}$$

Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.

Example $$\PageIndex{2}$$

A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?

Given: mass of metal, time, and efficiency

Strategy:

1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.

Solution:

A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:

$$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$$

B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.

C Using the definition of the faraday,

coulombs = (1.85 × 10−2mol e)(96,486 C/mol e) = 1.78 × 103 C / mole

The current in amperes needed to deliver this amount of charge in 12.0 h is therefore

\begin{align}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align}

Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.

Exercise $$\PageIndex{2}$$

A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?