Electrochemistry

Question 1

What chemical process is happening to the Ag atom in this reaction?

Ag(s) ⇒ Ag⁺(aq) + e^-

Answer 1

oxidation

The mnemonic to use in redox (reduction-oxidation) reactions is OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). In this case, the Ag atom is losing an electron, so it is being oxidized.

Question 2

What chemical process is happening to the Fe³⁺ ion in this reaction?

Fe³⁺(aq) + e^- ⇒ Fe²⁺(aq)

Answer 2

reduction

The mnemonic to use in redox (reduction-oxidation) reactions is OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). In this case, the Fe³⁺ ion is gaining an electron, so it is being reduced.

Question 3

Define:

electrochemical cell

Answer 3

It is a pair of chemical systems which either:

1. undergo a reaction and generate electric current
2. undergo a reaction when electric current is run through the system

Question 4

Identify the principle components of the electrochemical cell shown below.

Answer 4

It has two half-cells, each of which contains an electrolyte and an electrode.

The electrodes are connected via electrically conductive material (often a wire).

The half-cells can be separate, as below, or share an electrolyte, but there will always be two separate electrodes.

Question 5

What is the anode of an electrochemical cell?

Answer 5

It is the electrode of the half-cell in which oxidation is taking place.

The mnemonic for remembering that oxidation occurs at the anode is AN OX, RED CAT.

Question 6

What is the cathode of an electrochemical cell?

Answer 6

It is the electrode of the half-cell in which reduction is taking place.

The mnemonic for remembering that reduction occurs at the cathode is AN OX, RED CAT.

Question 7

Define:

electrolysis

Answer 7

It is the act of using a voltage source to drive a non-spontaneous electrochemical reaction.

Electrochemical cells in which electrolysis is occurring are referred to as electrolytic cells.

Question 8

In an electrochemical cell, at which electrode does oxidation occur? At which electrode does reduction occur?

Answer 8

• Oxidation occurs at the anode
• Reduction occurs at the cathode

A simple mnemonic to remember this by is AN OX, RED CAT. Here, ANode = OXidation, REDuction = CAThode.

Question 9

In an electrochemical cell, in which direction do electrons flow?

Answer 9

anode to cathode

Remember that REDuction occurs at the CAThode (RED CAT), and reduction Is the gain of electrons (OIL RIG), so electrons must be flowing to the cathode to facilitate reduction.

Question 10

What type of electrochemical cell is being depicted below?

Answer 10

electrolytic cell

The two main types of electrochemical cells (electrolytic and galvanic) can be easily discerned by the presence or absence of a power source. In this case, the battery is the power source, indicating an electrolytic cell.

Question 11

In the electrolytic cell depicted below, which electrode is the anode?

Answer 11

Electrode A

The positive end of the battery attracts electrons towards itself, and those electrons must be leaving Electrode A. Since electrons flow from anode to cathode, Electrode A must be the anode.

Note that in an electrolytic cell, the anode is the positive (+) electrode.

Question 12

In the electrolytic cell depicted below, which electrode is the cathode?

Answer 12

Electrode B

The negative end of the battery repels electrons away from itself, and those electrons will move toward Electrode B. Since electrons flow from anode to cathode, Electrode B must be the cathode.

Note that in an electrolytic cell, the cathode is the negative (-) electrode.

Question 13

In the electrolysis of water shown below, which gaseous species will appear at the anode?

Answer 13

O₂(g)

The anode is where oxidation occurs, so look for a species which can lose electrons.

Water can be thought of as consisting of O^2- ions and H⁺ ions. The O^2- ions can lose electrons to form molecular oxygen, O₂, and this will occur at the anode.

Question 14

In the electrolysis of NaCl shown below, which chemical species will appear at the anode?

Answer 14

Cl₂(g)

NaCl is made up of Na⁺ and Cl^- ions. Cl^- ions will lose electrons at the anode to form molecular chlorine, Cl₂.

Question 15

In the electrolysis of water shown below, which gaseous species will appear at the cathode?

Answer 15

H₂(g)

The cathode is where reduction occurs, so look for a species which can gain electrons.

Water can be thought of as consisting of O^2- ions and H⁺ ions. The H⁺ ions can gain electrons to form molecular hydrogen, H₂, and this will occur at the cathode.

Question 16

In the electrolysis of NaCl shown below, which chemical species will appear at the cathode?

Answer 16

Na(s)

NaCl is made up of Na⁺ and Cl^- ions. Na⁺ ions will gain electrons at the cathode to form sodium metal.

Question 17

What is an electrolyte, in the context of an electrochemical cell?

Answer 17

It is any species added to the cell (in which an electrochemical reaction is occurring) that allows current to flow more easily.

For example, in the electrolysis of water, either acid or salt must be added to the water to improve current flow. Pure water is not a good conductor of electricity.

Question 18

What are the units of current (I)?

Answer 18

amperes

(A)

Current is defined as charge over time, or Q/t. Therefore, 1 A = 1 C/s.

Question 19

If 2.0 A of current flows for three minutes, how many C of charge flow in this time?

Answer 19

360 C

I = 2.0 A = 2.0 C/s

Q = I x t = 2.0 * 3 min * 60s/min = 360 C

Question 20

What equation relates current to number of moles of electrons?

Answer 20

Faraday’s equation, I x t = n x F

where

• I = current (A)
• t = time the current flows (s)
• n = # of moles of electrons
• F = Faraday’s constant, 96,485 C/mol e^-

For the MCAT, approximate F = 10⁵ C/mol e^-.

Question 21

In the electrolytic cell shown below, if a current of 1.0 A flows for 10 seconds, how many grams of Na(s) are plated onto the cathode?

Answer 21

2.3 x 10^-3 g of Na will be deposited.

Faraday’s equation can be used to calculate the number of moles of electrons which flow through the circuit:

n = I x T / F
1.0 x 10 / 10⁵ = 1 x 10^-4 mol e^-

Each mole of electrons reduces one mole of Na⁺. The mass of 10^-4 mol Na(s) is:

m = 1x10^-4 * 23g/mol
2.3x10^-3 g Na

Question 22

Which type of electrochemical cell is being depicted below?

Answer 22

galvanic cell

The two main types of electrochemical cells (electrolytic and galvanic) can be easily discerned by the presence or absence of a power source. In this case, the lack of a battery indicates that this is a galvanic cell.

Question 23

In the galvanic cell depicted below, which electrode is the anode?

Answer 23

The Zn electrode

In any electrochemical cell, electrons flow from anode to cathode. Electrons are leaving the Zn electrode, so it must be the anode.

Question 24

In the galvanic cell depicted below, which electrode is the cathode?

Answer 24

The Cu electrode

In any electrochemical cell, electrons flow from anode to cathode. Electrons are arriving at the Cu electrode, so it must be the cathode.

Question 25

What is the charge on the cathode of a galvanic cell?

Answer 25

positive (+)

The cathode spontaneously attracts electrons, which is what would be expected of a positively-charged electrode. This is the opposite convention from electrolytic cells, where the cathode has a negative charge.

Question 26

What is the charge on the anode of a galvanic cell?

Answer 26

negative (-)

The anode spontaneously discharges electrons, which is what would be expected of a negatively-charged electrode. This is the opposite convention from electrolytic cells, where the anode has a positive charge.

Question 27

In the galvanic cell depicted below, will the Zn anode become lighter or heavier as current flows through the circuit?

Answer 27

lighter

Oxidation occurs at the anode in any electrochemical cell. In this case, the oxidation reaction will result in Zn(s) becoming Zn²⁺(aq). As the Zn ions move into solution, the Zn electrode will get lighter.

Question 28

In the galvanic cell depicted below, will the Cu cathode become lighter or heavier as current flows through the circuit?

Answer 28

heavier

Reduction occurs at the cathode in any electrochemical cell. In this case, the reduction reaction will result in Cu²⁺(aq) becoming Cu(s). As the Cu²⁺ ions move out of solution, they will plate onto the Cu electrode, which will get heavier.

Question 29

In the galvanic cell depicted below, which electrode will metal plate onto?

Answer 29

Cu ions will plate onto the surface of the cathode, which will get heavier.

Question 30

If the overall reaction of an electrochemical cell is:

Co³⁺(aq) + Na(s) ⇒ Co²⁺(aq) + Na⁺(aq)

what is the oxidation half-reaction?

Answer 30

Na(s) ⇒ Na⁺(aq) + e^-

An oxidation half-reaction follows only the species that gets oxidized in a redox rection, as well as the electron or electrons involved.

Question 31

If the overall reaction of an electrochemical cell is:

Co³⁺(aq) + Na(s) ⇒ Co²⁺(aq) + Na⁺(aq)

what is the reduction half-reaction?

Answer 31

Co³⁺(aq) + e^- ⇒ Co²⁺(aq)

A reduction half-reaction follows only the species that gets reduced in a redox reaction, as well as the electron or electrons involved.

Question 32

Define:

reduction potential

(Eº_red)

Answer 32

It is the amount of voltage either required or liberated by the reduction half-reaction.

The more positive a species’ reduction potential, the more strongly it will tend to be reduced in electrochemical cells. Redox potentials are usually written as reduction potentials, but on the MCAT, make sure to check.

Question 33

If the potential of the half-reaction

Co³⁺(aq) + e^- ⇒ Co²⁺(aq)

is Eº_red = +1.82V, what is the potential of a cell which is reduces Co³⁺ to Co²⁺?

Answer 33

Eº_red = +1.82 V

In general, the reduction potential gives the amount of voltage released or required by a cell for the reduction reaction to occur. Since the Eº_red > 0 for this cell, it is favorable to reduce Co³⁺; in other words, energy is released when this process happens.

Question 34

If the potential of the half reaction

Co³⁺(aq) + e^- ⇒ Co²⁺(aq)

is Eº_red = +1.82V, what is the potential of a cell which oxidizes Co²⁺ to Co³⁺?

Answer 34

Eº_ox = -1.82 V

An oxidation reaction is the reverse of a reduction reaction, in this case, the reverse of the reduction of Co³⁺. The oxidation potential Eº_ox is simply the reverse of Eº_red. Since Eº_ox < 0, the oxidation is unfavorable unless 1.82 V are supplied to oxidize this system.

Question 35

If, in an electrochemical cell, the potential of the oxidation half-reaction is Eº_ox and the potential of the reduction half-reaction is Eº_red, what is the full potential of the cell?

Answer 35

Eº_cell = Eº_ox + Eº_red

Remember to keep track of the appropriate signs of the oxidation and reduction half-reactions; a classic MCAT trick will give you a reduction potential, then ask about the oxidation potential for the same reaction.

Question 36

What is the sign of the cell potential for a galvanic cell?

Answer 36

Eº_cell > 0

Since a galvanic cell is spontaneous, it must have a favorable voltage, which means it is positive in sign.

Question 37

What is the sign of the cell potential for an electrolytic cell?

Answer 37

Eº_cell < 0

Since an electrolytic cell is non-spontaneous, it must have an unfavorable voltage, which means it is negative in sign.

Question 38

If the following two reactions are combined into a single galvanic cell, what is the cell’s total potential?

Co³⁺(aq) + e^- ⇒ Co²⁺(aq) Eº = 1.82 V
Na⁺(aq) + e^- ⇒ Na(s) Eº = -2.71 V

Answer 38

+4.53 V

Since the cell is galvanic, its potential must be positive. So the two reactions must be combined (one oxidation, one reduction) to yield an overall positive cell potential. In this case, Co must be reduced (Eº_red = +1.82V) while Na is oxidized (Eº_ox = -(-2.71V)).

So the total potential is:

Eº_cell = Eº_red(Co) + Eº_ox(Na)
= 1.82 + 2.71 = 4.53 V

Question 39

If the following two reactions are combined into a single electrolytic cell, what is the cell’s total potential?

Co³⁺(aq) + e^- ⇒ Co²⁺(aq) Eº = 1.82 V
Na⁺(aq) + e^- ⇒ Na(s) Eº = -2.71 V

Answer 39

-4.53 V

Note that the answer is exactly the reverse of these two cells arranged as a galvanic cell. This must be true, since the electrolytic cell is just the galvanic cell run backwards. Here, Co is being oxidized while Na is being reduced, and the total potential is:

Eº_cell = Eº_red(Na) + Eº_ox(Co)
= -2.71 + (-1.82) = -4.53 V

Question 40

What equation relates an electrochemical cell’s standard potential Eº to its standard Gibbs’ free energy ΔGº?

Answer 40

ΔGº = -nFEº

Where:

n = number of moles of electrons in the equation
F = 10⁵ C/mol electrons

Question 41

According to the equation

ΔGº = -nFEº

what is the sign of ΔGº for an electrolytic cell?

Answer 41

ΔGº > 0

The standard voltage (Eº) of an electrolytic cell is negative, which means ΔGº must be positive. This can also be derived from the fact that an electrolytic cell is non-spontaneous, and any non-spontaneous reaction has a positive ΔGº.

Question 42

According to the equation

ΔGº = -nFEº

what is the sign of ΔGº for a galvanic cell?

Answer 42

ΔGº < 0

The standard voltage (Eº) of a galvanic cell is positive, which means ΔGº must be negative. This can also be derived from the fact that a galvanic cell is spontaneous, and any spontaneous reaction has a negative ΔGº.

Question 43

What equation relates the standard voltage of an electrochemical cell (Eº) to its equilibrium constant (K_eq)?

Answer 43

nFEº = RT ln(K_eq)

Where:

R = ideal gas constant (J/K*mol)
T = temperature (K)
n = number of moles of electrons
F = 10⁵ C/mol electrons

Question 44

According to the equation

nFEº = RT ln(K_eq)

what is the value of K_eq for a galvanic cell?

Answer 44

K_eq > 1

For a galvanic cell, Eº is positive, so ln(K_eq) must be positive, meaning that k_eq > 1. This can also be derived from the fact that a galvanic cell is spontaneous, and K_eq > 1 (it favors the products) for any spontaneous reaction.

Question 45

According to the equation

nFEº = RT ln(K_eq)

what is the value of K_eq for an electrolytic cell?

Answer 45

K_eq < 1

For an electrolytic cell, Eº is negative, so ln(k_eq) must be negative, meaning that 0 < K_eq < 1. This can also be derived from the fact that an electrolytic cell is non-spontaneous, and 0 < K_eq < 1 (it favors the reactants) for any non-spontaneous reaction.

Question 46

What equation can be used to calculate the voltage of an electrochemical cell which is not at standard conditions?

Answer 46

nFE = -RT ln(Q/K_eq)

This is similar to the equation for calculating the standard voltage. In fact, at standard conditions (where Q = 1), the two equations are the same.

Question 47

According to the equation

nFE = -RT ln(Q/K_eq)

what is the voltage of a cell for which Q < K_eq?

Answer 47

E_cell > 0

If Q < K_eq, the reaction has not yet reached equilibrium; it must move forward to reach it. For this reason, the reaction will proceed spontaneously, the cell is galvanic, and E_cell is positive.

Question 48

According to the equation

nFE = -RT ln(Q/K_eq)

what is the voltage of a cell for which Q > K_eq?

Answer 48

E_cell < 0

If Q > K_eq, the reaction has exceeded equilibrium and must proceed backwards to reach it. Since E_cell always refers to the forward reaction, E_cell is negative in this case.

Question 49

According to the equation

nFE = -RT ln(Q/K_eq)

what is the voltage of a cell for which Q = K_eq?

Answer 49

E_cell = 0

If Q = K_eq, the reaction is at equilibrium, and will remain at its current state. Since neither the forward nor backward reaction will proceed, E_cell = 0.