Quantitative Aspects of Electrolysis

Converting Current to Moles of Electrons

In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?

Solution

Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time

\(n=\phantom{\rule{0.2em}{0ex}}\cfrac{Q}{F}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\cfrac{\text{10.23 C}}{\text{s}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{1 hr}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{60 min}}{\text{hr}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{60 s}}{\text{min}}}{9{\text{6,485 C/mol e}}^{\text{−}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{36,830 C}}{\text{96,485 C/mol}\phantom{\rule{0.2em}{0ex}}{\text{e}}^{\text{−}}}\phantom{\rule{0.2em}{0ex}}=0.381{\text{7 mol e}}^{\text{−}}\)

From the problem, the solution contains AgNO₃, so the reaction at the cathode involves 1 mole of electrons for each mole of silver

\({\text{cathode: Ag}}^{\text{+}}(aq)+{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Ag}(s)\)

The atomic mass of silver is 107.9 g/mol, so

\(\text{mass Ag}={\text{0.3817 mol e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 mol Ag}}{{\text{1 mol e}}^{\text{−}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{107.9 g Ag}}{\text{1 mol Ag}}\phantom{\rule{0.2em}{0ex}}=4\text{1.19 g Ag}\)

Check your answer: From the stoichiometry, 1 mole of electrons would produce 1 mole of silver. Less than one-half a mole of electrons was involved and less than one-half a mole of silver was produced.

Time Required for Deposition

In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m² from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm³.

Solution

This problem brings in a number of topics covered earlier. An outline of what needs to be done is:

• If the total charge can be determined, the time required is just the charge divided by the current
• The total charge can be obtained from the amount of Cr needed and the stoichiometry
• The amount of Cr can be obtained using the density and the volume Cr required
• The volume Cr required is the thickness times the area

Solving in steps, and taking care with the units, the volume of Cr required is

\(\text{volume}=\phantom{\rule{0.1em}{0ex}}(\text{0.010 mm}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 cm}}{\text{10 mm}})\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}(3.3{\phantom{\rule{0.2em}{0ex}}\text{m}}^{2}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}(\cfrac{10,000{\phantom{\rule{0.2em}{0ex}}\text{cm}}^{2}}{1{\phantom{\rule{0.2em}{0ex}}\text{m}}^{2}}))\phantom{\rule{0.1em}{0ex}}={\text{33 cm}}^{3}\)

Cubic centimeters were used because they match the volume unit used for the density. The amount of Cr is then

\(\text{mass}=\text{volume}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{density}=33\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{{\text{cm}}^{3}}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{7.19 g}}{\require{cancel}\cancel{{\text{cm}}^{3}}}\phantom{\rule{0.2em}{0ex}}=\text{237 g Cr}\)

\(\text{mol Cr}=\text{237 g Cr}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 mol Cr}}{\text{52.00 g Cr}}\phantom{\rule{0.2em}{0ex}}=\text{4.56 mol Cr}\)

Since the solution contains chromium(III) ions, 3 moles of electrons are required per mole of Cr. The total charge is then

\(Q=\text{4.56 mol Cr}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{3{\phantom{\rule{0.2em}{0ex}}\text{mol e}}^{\text{−}}}{\text{1 mol Cr}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{96485 C}}{{\text{mol e}}^{\text{−}}}\phantom{\rule{0.3em}{0ex}}=1.32\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{C}\)

The time required is then

\(t=\phantom{\rule{0.2em}{0ex}}\cfrac{Q}{I}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{1.32\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{C}}{\text{33.46 C/s}}\phantom{\rule{0.2em}{0ex}}=3.95\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{s}=\text{11.0 hr}\)

Check your answer: In a long problem like this, a single check is probably not enough. Each of the steps gives a reasonable number, so things are probably correct. Pay careful attention to unit conversions and the stoichiometry.