AP Topic 15: Electrochemistry

Goes with chapter 21: Silberberg’s Principles of General ChemistryMrs. Laura Peck, 2013

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Objectives/Study guide Identify and compare the two types of electrochemical cells:

galvanic and electrolytic Draw and label a galvanic cell, including labeling the electrodes,

the flow of electrons, and the flow of ions Write half-reactions and determine which reaction occurs at the

anode and which reaction occurs at the cathode. Give the line notation for a galvanic cell or write a balanced redox

reaction from the given line notation. Calculate the cell potential for a galvanic cell and an electrolytic

cell. Determine if a reaction is spontaneous from its cell potential. Calculate the cell potential under nonstandard conditions when

the solutions are not 1M. This involves the use of the Nernst equation.

Determine the strengths of oxidizing agents and reducing agents. Draw and label an electrolytic cell. Determine the reactions which occur at the anode and the

cathode during electrolysis. Perform stoichiometric calculations involving electrolysis.

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Basic differences in cells. Electrochemistry is the study of the interchange of

electrical and chemical energy. There are two types of electrochemical cells,

galvanic and electrolytic. In galvanic cells, spontaneous redox reactions generate

electric current. In electrolytic cells, a nonspontaneous chemical reaction

occurs with the application of an electric current.

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Line Notation A galvanic cell can be abbreviated with line

notation. Reactant/product II reactant/product (anode reaction) (cathode reaction)

The salt bridge is indicated by the symbol II

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Example #1: Give the correct line notation for the Galvanic cell pictured.

Zn/Zn2+ II Cu2+/Cu

AP tips: Here are some mnemonic devices to help you remember

some facts about electrochemistry and redox reactions. ‘LEO’ goes ‘GER’ means Loss of Electrons is Oxidation

and Gain of Electrons is Reduction. To recall what happens at the anode and the cathode:

RedCat and AnOx means Reduction Occurs at the cathode and oxidation occurs at the anode.

To know the migration of ions toward the electrodes for both types of cells, ‘Cat’ ions move to the ‘Cat’ode and ‘An’ ions move to the ‘An’ode.

You will be provided with a table of standard reduction potentials on the AP test.

You should be able to sketch a galvanic cell and label the electrodes, the flow of electrons, and the flow of ions.

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Standard reduction potentials The cell potential, Ecell, is the potential of the cell to do work on

its surroundings by driving an electric current through a wire. By definition, a potential of 1 volt is produced when 1 joule of energy

moves 1 coulomb of electric charge across a potential.

The magnitude of the cell potential is a measure of the driving force behind an electrochemical reaction. Sometimes it is referred to as the electromotive force or emf.

Tables of reduction potentials give standard voltages for reduction half-reactions measured at standard conditions of 1 atm, 1 molar solution, and 25*C

The reaction occurring in a galvanic cell can be broken down into an oxidation half-reaction and a reduction half-reaction.

Using the table of standard reduction potentials in your text, you can calculate the cell potential of the overall reaction.

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Example #2: consider a galvanic cell based on the reaction: Al + NI2+ Al3+ + Ni

Give the balanced cell reaction and calculate the cell potential, E0

cell For the reaction.

Step 1: Write the oxidation & reductionHalf-reactions.

Step 2: For the reduction half-reactionLook up the potential in your book.

Step 3: For the oxidation half-reaction, E0

ox = -E0red

Step 4: the cell potential for the overallReaction is equal to the sum of theReduction potential, E0

red, and the Oxidation potential, E0

ox.

Step 5: to obtain the balanced cellReaction, you must make sure that theElectrons lost equal the electrons gained. When multiplying the half-reactions throughBy a coefficient, do not change the value of E0

Oxidation: Al Al3+ + 3e-Reduction: Ni2+ + 2e- Ni

Ni2+ + 2e- Ni E0red = -0.23V

Oxidation: Al Al3+ + 3e- E0

ox= -E0red -(-1.66V) = +1.66V

E0cell= E0

ox + E0red

E0cell = -0.23V + 1.66V = 1.43V

3[Ni2+ + 2e- Ni] E0red = -0.23V

2[Al Al3+ + 3e-] E0ox = +1.66V

3Ni2+ + 2Al 3Ni + 2Al3+ E0cell = 1.43V

Spontaneous Reactions Gibbs free energy, ΔG°, can be calculated

from the cell potential, E0cell.

ΔG° = -nFE0cell

Faraday’s constant, F, has a value of

96,485 C/mol e- The number of moles of electrons

transferred in a redox reaction is represented by n A spontaneous reaction is one that has a negative value for ΔG°

or a positive value for E0cell

You may be asked if an element or ionic species is capable of reducing another element or ion.

To determine if the reaction will occur, write the half-reactions and calculate the cell potential .

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Example #3: Will 1M HCl dissolve silver metal and form Ag+ solution?

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Write half-reactions and Calculate E0

cell

2H+ + 2e- 2H2 E0red = 0.00V

2Ag 2Ag+ + 2e- E0ox = -0.80V

2H+ + 2Ag H2 + 2Ag+ E0cell = -0.80V

The negative value for E0cell indicates that

the reaction will not occur.

Example #4: Bromine, Br2, can oxidize iodide, I-, to iodine, I2. However, Br2 cannot oxidize chloride, Cl-, to chlorine, Cl2. Explain why the first reaction occurs yet, the second one does not.

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Begin by writing the appropriateHalf-reactions. Then calculateThe cell potential for the overallReaction.

First, the reaction in which Br2 Oxidizes I-

Then the reaction in which Br2

Oxidizes Cl-

Br2 + 2e- 2Br- E0red = 1.09V

2I- I2 + 2e- E0ox = -0.54V

Br2 + 2I- 2Br- + I2 E0cell = 0.55V

This reaction occurs, E0cell is +

Br2 + 2e- 2Br- E0red = 1.09V

2Cl- Cl2 + 2e- E0ox = -1.36V

Br2 + 2Cl- 2Br- + Cl2 E0cell = -0.27V

This reaction does not occur, E0cell is negative

Cell Dependence on Concentration The galvanic cell represented by the reaction:

3Ni2+ + 2Al 3Ni + 2Al3+

Has a cell potential, E0cell, equal to 1.43V under

standard conditions (all solutions are 1M) Increasing the concentration of Ni2+ will shift the

reaction to the right by Le Chatelier’s principle, increasing the driving force on the electrons and increasing the cell potential.

The relationship between the cell potential and concentrations at 25°C is given by the Nernst equation: Ecell = E0

cell – (0.0591/n)log Q

The cell potential, Ecell, is for nonstandard conditions. The moles of electrons transferred are represented by n The mass action quotient is represented by Q

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Example #5: Calculate the cell potential for the reaction: 3Ni2+ + 2Al 3Ni + 2Al3+

In which [Al3+] = 2.00M and [Ni2+] = 0.750M (you already know the E0

cell = 1.43V and the number of moles of electrons transferred, n, equals 6)

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Q = [Al3+]2 / [Ni2+]3 = (2.00)2 / (0.750)3 = 9.48

Ecell = 1.43V – (0.0591/6) log 9.48 = 1.33V

Determining the Strength of Oxidizing and Reducing Agents.

You may be asked to list atoms or ions in order of increasing strength as reducing agents or oxidizing agents.

For a substance to be oxidized, it must lose electrons and another substance must gain electrons because oxidation and reduction always occur together.

The substance that causes another substance to be oxidized is called an oxidizing agent.

An oxidizing agent is reduced; it is the reactant in the reduction half-reaction.

The larger (more positive) E0red, the stronger the oxidizing

agent. A reducing agent is oxidized; it is the reactant in the oxidation

half-reaction. The larger (more positive) the E0

ox, the stronger the reducing agent.

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Example #6: classify each of the following as an oxidizing agent, reducing agent, or both. Within each list, arrange in order of increasing strength as oxidizing agents and reducing agents.

Br2, Mg, Fe2+, I2, Cl-, Cu2+

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To be an oxidizing agent, a substance must be capable of gainingElectrons or being reduced. Of the species listed, Mg and Cl- are The only ones listed that cannot have a lower oxidation state. ForThe oxidizing agents listed above, the respective reduction potentialsAre -0.44V, 0.16V, or 0.34V for Cu2+ (which can be reduced to Cu0

Or Cu+), 0.54V, and 1.09V. The more positive the cell potential, theStronger the oxidizing agent.

Reducing agents must be capable of being oxidized to a higherOxidation state. Cl- and Mg can go the Cl0 and Mg2+. Fe2+ can Exist as Fe3+ or Fe0 so it can act as an oxidizing agent or reducingAgent. For the reducing agents listed above, the correspondingOxidation potentials are -1.36V, -0.77V, + 2.37V. The more positiveThe cell potential, the stronger the reducing agent.

Oxidizing: Fe2+<Cu2+<I2<Br2

Reducing: Cl-<Fe2+<Mg

Electrolytic cells In an electrolytic cell, a nonspontaneous reaction is made to occur by

forcing an electric current through the cell. In an earlier example, it was shown that the following reaction is

spontaneous: 3Ni2+ + 2Al 3Ni + 2Al3+

The reverse of this reaction: 3Ni + 2Al3+ 3Ni2+ + 2Al is nonspontaneous and can be made to occur by the addition of an external power source.

This electrolytic cell can be set up with two compartments just like the galvanic cell, with the replacement of a power supply for the voltmeter.

In the process of electroplating, the electrolytic cell can also be set up using only one compartment

For example, if an object is to be plated with copper, make it the cathode and immerse it into a copper(II)sulfate solution.

At the cathode, the reaction that will occur and deposit copper onto the object is Cu2+ + 2e- Cu.

The anode can also be made of copper. The oxidation of copper occurs at this anode.

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Reactions that occur in an electrolytic cell To determine which

reaction occurs at the anode and the cathode during electrolysis, you must consider all possible reactions and their reduction and oxidation potentials.

If the reaction takes place in an aqueous solution, the oxidation and reduction of water must be considered.

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Example #7: A solution of copper(II)sulfate is electrolyzed. Calculate the cell potential of the reaction, E0

cell.

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possible reactions Cell potential, Eo(V)Cathode Cu2+ + 2e- Cu 0.34 SO4

2- + 4H+ + 2e- H2SO3 + H2O 0.20 2H2O + 2e- H2 + 2OH- -0.83

anode Cu Cu2+ + 2e- -0.34 2H2O O2 + 4H+ + 4e- -1.23

For each electrode, the reaction with the more positive potentialWill occur. At the cathode, Cu2+ will be reduced. At the anode, CuWill be oxidized.

Cu2+ + 2e- Cu Cu Cu2+ +2e-

AP Tip

Frequently, the electrodes are inert for electrolysis.

For example, during the electrolysis of KI(aq) - K+, I-, and H2O are the only species present.

Only I- and H2O are present to be oxidized at the anode.

Note: In aqueous KI, there is no K(s) to be oxidized.

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Stoichiometry of electrolytic processes. Lets review how much chemical change

occurs with the flow of a given current for a specified time.

You might be asked how much metal was plated (formed) or how long an electroplating process will take or how much current is required to produce a specified amount of metal over a period of time.

Some units to be familiar with include A, amperes; 1A = 1C/s; coulombs, C; Faraday’s constant is 96,4895 C = 1mol e-

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Example #8: A current of 10.0A is passed through a solution containing M2+

for 30.0 min. It produces 5.94 g of metal, M. Determine the identity of metal, M.

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10.0A = 10.0C/s x 30.0 min x 60 s/min = 1.80x104 C

1.80x104 C x 1 mol e- /96,485C x 1 mol M/2 mol e- = 9.33x10-2 mol M

Molar mass of M = g M/mol M = 5.94gM / 9.33x10-2 mol M = 63.7g/mol

63.7 g/mol is the molar mass of Cu. You can also do this is one step

5.94g M x 2 mol e- x 96485C x 1 s x 1 min x 1 = 63.7g/mol 1 mol M 10.0C mole e- 60 s 30.0 min

Comparison of Galvanic and Electrolytic cells

Galvanic and electrolytic cells have few features in common.

For both types of cells, reduction always occurs at the cathode and oxidation at the anode.

In an electrolytic cell, electrons travel from the battery to the cathode.

In both cases, electrons travel in the wire, but you wouldn’t say the electrons travel from the anode to the cathode in an electrolytic cell.

Positive ions are always attracted to the cathode whether the cell is electrolytic or galvanic.

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Comparison cont…Galvanic Electrolytic

Sign of the cathode + -

Sign of the anode - +

Ions attracted to the cathode

Cations (+) Cations (+)

Ions attracted to the anode

Anions (-) Anions (-)

Sign of E0cell + -

Spontaneity Spontaneous nonspontaneous

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In both types of cells, the + ions or cations move toward the cathodeBecause there is an excess of negative ions at the cathode caused byThe reduction of + ions in solution.

Likewise, oxidation at the anode produces + ions, so negative ions or anions in the salt bridge must move to the anode to maintain electrical Neutrality.

The end…

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