# 10.6 Electrolysis  (Page 3/7)

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$Q=I\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}t=n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}F$

Where t is the time in seconds, n the number of moles of electrons, and F is the Faraday constant.

Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.

## 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

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

From the problem, the solution contains AgNO 3 , so the reaction at the cathode involves 1 mole of electrons for each mole of silver

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

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

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

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.

Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 $×$ 10 3 A passed through the solution for 15.0 minutes? Assume the yield is 100%.

${\text{Al}}^{3+}\left(aq\right)+3\phantom{\rule{0.2em}{0ex}}{\text{e}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Al}\left(s\right);$ 7.77 mol Al = 210.0 g Al.

## 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 m 2 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/cm 3 .

## 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

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

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

$\text{mass}=\text{volume}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{density}=33\phantom{\rule{0.2em}{0ex}}\overline{){\text{cm}}^{3}}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{7.19 g}}{\overline{){\text{cm}}^{3}}}\phantom{\rule{0.2em}{0ex}}=\text{237 g Cr}$
$\text{mol Cr}=\text{237 g Cr}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\frac{\text{1 mol Cr}}{\text{52.00 g Cr}}\phantom{\rule{0.2em}{0ex}}=\text{4.56 mol Cr}$

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

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

The time required is then

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

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.

What mass of zinc is required to galvanize the top of a 3.00 m $×$ 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO 3 ) 2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm 3 .

231 g Zn required 446 minutes.

## Key concepts and summary

Using electricity to force a nonspontaneous process to occur is electrolysis. Electrolytic cells are electrochemical cells with negative cell potentials (meaning a positive Gibbs free energy), and so are nonspontaneous. Electrolysis can occur in electrolytic cells by introducing a power supply, which supplies the energy to force the electrons to flow in the nonspontaneous direction. Electrolysis is done in solutions, which contain enough ions so current can flow. If the solution contains only one material, like the electrolysis of molten sodium chloride, it is a simple matter to determine what is oxidized and what is reduced. In more complicated systems, like the electrolysis of aqueous sodium chloride, more than one species can be oxidized or reduced and the standard reduction potentials are used to determine the most likely oxidation (the half-reaction with the largest [most positive] standard reduction potential) and reduction (the half-reaction with the smallest [least positive]standard reduction potential). Sometimes unexpected half-reactions occur because of overpotential. Overpotential is the difference between the theoretical half-reaction reduction potential and the actual voltage required. When present, the applied potential must be increased, making it possible for a different reaction to occur in the electrolytic cell. The total charge, Q , that passes through an electrolytic cell can be expressed as the current ( I ) multiplied by time ( Q = It ) or as the moles of electrons ( n ) multiplied by Faraday’s constant ( Q = nF ). These relationships can be used to determine things like the amount of material used or generated during electrolysis, how long the reaction must proceed, or what value of the current is required.

## Key equations

• Q = I $×$ t = n $×$ F

## Chemistry end of chapter exercises

Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Appendix L are the same as those at each of the melting points. Assume the efficiency is 100%.

(a) CaCl 2

(b) LiH

(c) AlCl 3

(d) CrBr 3

What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 $×$ 10 5 C passes through each cell? Assume the voltage is sufficient to perform the reduction.

(a) $\begin{array}{}\\ \text{mass Ca}=\text{69.1 g}\\ {\text{mass Cl}}_{2}=\text{122 g}\end{array};$ (b) $\begin{array}{l}\text{mass Li}=\text{23.9 g}\\ {\text{mass H}}_{2}=\text{3.48 g}\end{array};$ (c) $\begin{array}{l}\text{mass Al}=\text{31.0 g}\\ {\text{mass Cl}}_{2}=\text{122 g}\end{array};$ (d) $\begin{array}{l}\text{mass Cr}=\text{59.8 g}\\ {\text{mass Br}}_{2}=\text{276 g}\end{array}$

How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction.

(a) Al 3+ , 1.234 A

(b) Ca 2+ , 22.2 A

(c) Cr 5+ , 37.45 A

(d) Au 3+ , 3.57 A

A current of 2.345 A passes through the cell shown in [link] for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)

0.79 L

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO 3 ) 2 solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm 3 . Assume the efficiency is 100%.

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