Electrochemistry

Question 1

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

Ag(s) ⇒ Ag⁺(aq) + e^-

Answer 1

The Ag is undergoing 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 is losing an electron, so it is being oxidized.

Question 2

What is the chemical process happening to Fe in this reaction?

Fe³⁺(aq) + e^- ⇒ Fe²⁺(aq)

Answer 2

The Fe³⁺ is undergoing 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⁺ is gaining an electron, so it is being reduced.

Question 3

Define:

An electrochemical cell

Answer 3

An electrochemical cell is a set of chemical systems which either:

1. undergo a reaction and generate electric current, or
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

An electrochemical cell has two half-cells, each of which has an electrolyte and an electrode. The electrodes are connected electrically.

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

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

Question 6

What is the cathode of an electrochemical cell?

Answer 6

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

Question 7

Define:

Electrolysis

Answer 7

Electrolysis is the act of using a voltage source to drive a non-spontaneous chemical reaction.

Electrochemical cells in which electrolysis is occuring 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

In any electrochemical cell, oxidation occurs at the anode, while reduction occurs at the cathode.

A simple mnemonic to remember this by:

AN OX, RED CAT

OXidation = ANode, REDuction = CAThode.

Question 9

In an electrochemical cell, what direction do electrons flow?

Answer 9

Electrons always flow from Anode to Cathode.

Remember: Reduction occurs at the Cathode (RED CAT), and Reduction Is 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

The cell is an 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 is the anode.

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: Since the anode is attached to the positive end of the battery, 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 is the cathode.

The negative end of the battery rejects electrons away from itself, and those electrons must reach Electrode B. Since electrons flow from anode to cathode, Electrode B must be the cathode.

Note: Since the cathode is attached to the negative end of the battery, 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) will appear at the anode.

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) will appear at the anode.

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) will appear at the cathode.

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) will appear at the anode.

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

Question 17

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

Answer 17

An electrolyte is any species added to the cell (in which an electrochemical reaction is occuring) which allows current to flow more easily.

Ex: in the electrolysis of water, either acid or salt must be added to the water to improve current flow, since pure water is not a good conductor of electricity.

Question 18

What are the units of current, I?

Answer 18

The units of current are amperes, A.

Current is defined as charge over time Q/t, so 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 is the equation relating current to number of moles of electrons?

Answer 20

Faraday’s Equation,
I x t = n x F

where:

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

Note: 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

2.3x10^-3 g of Na will be deposited.

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

n = I x T / F
= (1)10/10⁵ = 1x10^-4 mol e^-

Each mole of electrons reduces one mole of Na^+. Calculate the mass of 10^-4 mol of Na(s)

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

Question 22

What type of electrochemical cell is being depicted below?

Answer 22

The cell is an 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 is the anode.

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 is the anode.

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

The cathode has a positive (+) charge in a galvanic cell.

The cathode spontaneously attracts electrons, which is what would be expected of a positively-charged electrode. Note that 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

The anode has a negative (-) charge in a galvanic cell.

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

Question 27

In the galvanic cell depicted below, is the Zn anode going to get lighter or heavier as current flows through the circuit?

Answer 27

The Zn anode will get lighter as current flows.

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

Question 28

In the galvanic cell depicted below, is the Cu cathode going to get lighter or heavier as current flows through the circuit?

Answer 28

The Cu cathode will get heavier as current flows.

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

Question 29

In the galvanic cell depicted below, which electrode will have metal plated onto it and get heavier?

Answer 29

The Cu cathode will get heavier as Cu ions plate onto its surface.

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

The oxidation half-reaction is:

Na(s) ⇒ Na⁺(aq) + e^-

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

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

The reduction half-reaction is:

Co³⁺(aq) + e^- ⇒ Co²⁺(aq)

A reduction half-reaction follows only the species that gets reduction in a redox rection, and includes the electron or electrons that flow, as well.

Question 32

Define:

The reduction potential of an electrochemical cell Eº_red

Answer 32

Eº_red 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.

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 reducing 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⁺, and energy is released when it 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 is oxidizing 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, so in this case 1.82 V are required to oxidize this system. Since Eº_ox < 0, the oxidation is unfavorable to occur.

Question 35

If the oxidation potential of a redox half-reaction is Eº_ox, and the reduction potential is Eº_red, what is the total potential of the cell?

Answer 35

The cell’s full potential is:

Eº_cell = Eº_ox + Eº_red

Remember to keep track of the appropriate signs of the oxidation and reduction half-reactions!

Question 36

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

Answer 36

Eº_cell > 0 for any galvanic cell.

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 potential of an electrolytic cell?

Answer 37

Eº_cell < 0 for any electrolytic cell.

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

The cell’s total potential is +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 have an overall positive cell potential. In this case, that means that 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

The cell’s total potential is -4.53 V.

Note that the answer is exactly the inverse of these two cells arranged as a galvanic cell. This must be, since the electrolytic cell is just the galvanic cell run backwards. So 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 is the equation relating an electrochemical cell’s standard potential Eº to its standard Gibbs’ Free Energy ΔGº?

Answer 40

ΔGº = -nFEº

where

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

Question 41

According to the equation

ΔGº = -nFEº

what is the ΔGº of an electrolytic cell?

Answer 41

For an electrolytic cell, 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 ΔGº of a galvanic cell?

Answer 42

For a galvanic cell, 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 is the equation relating the standard voltage of an electrochemical cell Eº to its equilibrium constant k_eq?

Answer 43

nFEº = RT ln(k_eq)

Question 44

According to the equation

nFEº = RT ln(k_eq)

what is the value of k_eq for a galvanic cell?

Answer 44

For a galvanic cell, k_eq > 1.

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

Question 45

According to the equation

nFEº = RT ln(k_eq)

what is the value of k_eq for an electrolytic cell?

Answer 45

For an electrolytic cell, k_eq < 1.

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

Question 46

How can the voltage be calculated for an electrochemical cell which is not at standard conditions?

Answer 46

nFE = -RT ln(Q/k_eq)

Note that this is similar to the equation for calculating the standard voltage. In fact, at standard conditions (where Q = 1), this equation becomes the calculation for standard voltage.

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

If Q < k_eq, E_cell > 0.

If Q < k_eq, the reaction has not yet reached equilibrium, and will move forwards to reach it. So, the reaction will proceed spontaneously, the cell is galvanic, and E_cell will be 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

If Q > k_eq, E_cell < 0.

If Q > k_eq, the reaction has exceeded equilibrium, and will proceed backwards to reach it. So, the reverse reaction will proceed spontaneously, the cell is electrolytic, and E_cell will be negative.

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

If Q = k_eq, E_cell = 0.

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