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MCAT Gen. Chemistry > Electrochemistry > Flashcards

Flashcards in Electrochemistry Deck (49)
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1
Q

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

Ag(s) ⇒ Ag+(aq) + e-

A

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

2
Q

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

Fe3+(aq) + e- ⇒ Fe2+(aq)

A

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

3
Q

Define:

electrochemical cell

A

An electrochemical cell is a pair 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.
4
Q

Identify the principle components of the electrochemical cell shown below.

A

An electrochemical cell 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.

5
Q

What is the anode of an electrochemical cell?

A

The anode 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.

6
Q

What is the cathode of an electrochemical cell?

A

The cathode 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.

7
Q

Define:

electrolysis

A

Electrolysis 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.

8
Q

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

A

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

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

9
Q

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

A

Electrons always flow from 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.

10
Q

What type of electrochemical cell is being depicted below?

A

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.

11
Q

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

A

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

12
Q

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

A

Electrode B is the cathode.

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.

13
Q

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

A

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 O2- ions and H+ ions. The O2- ions can lose electrons to form molecular oxygen, O2, and this will occur at the anode.

14
Q

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

A

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.

15
Q

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

A

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 O2- ions and H+ ions. The H+ ions can gain electrons to form molecular hydrogen, H2, and this will occur at the cathode.

16
Q

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

A

Na(s) will appear at the cathode.

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

17
Q

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

A

An electrolyte 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.

18
Q

What are the units of current (I)?

A

The units of current are amperes (A).

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

19
Q

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

A

360 C

I = 2.0 A = 2.0 C/s

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

20
Q

What equation relates current to number of moles of electrons?

A

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-.

21
Q

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?

A

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

22
Q

Which type of electrochemical cell is being depicted below?

A

The cell is a 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.

23
Q

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

A

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.

24
Q

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

A

The Cu electrode is the cathode.

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

25
Q

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

A

In a galvanic cell, the cathode has a positive (+) charge.

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.

26
Q

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

A

In a galvanic cell, the anode has a negative (-) charge.

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.

27
Q

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

A

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) becoming Zn2+(aq). As the Zn ions move into solution, the Zn electrode will get lighter.

28
Q

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

A

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) becoming Cu(s). As the Cu2+ ions move out of solution, they will plate onto the Cu electrode, which will get heavier.

29
Q

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

A

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

30
Q

If the overall reaction of an electrochemical cell is:

Co3+(aq) + Na(s) ⇒ Co2+(aq) + Na+(aq)

what is the oxidation half-reaction?

A

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, as well as the electron or electrons involved.

31
Q

If the overall reaction of an electrochemical cell is:

Co3+(aq) + Na(s) ⇒ Co2+(aq) + Na+(aq)

what is the reduction half-reaction?

A

The reduction half-reaction is:

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.

32
Q

Define:

reduction potential (red)

A

The reduction potential (Eºred) of an electrochemical cell 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.

33
Q

If the potential of the half-reaction

Co3+(aq) + e- ⇒ Co2+(aq)

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

A

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 Eºred > 0 for this cell, it is favorable to reduce Co3+; in other words, energy is released when this process happens.

34
Q

If the potential of the half reaction

Co3+(aq) + e- ⇒ Co2+(aq)

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

A

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 Eºox is simply the reverse of Eºred. Since Eºox < 0, the oxidation is unfavorable unless 1.82 V are supplied to oxidize this system.

35
Q

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

A

The cell’s full potential is:

cell = Eºox + Eºred

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.

36
Q

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

A

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.

37
Q

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

A

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.

38
Q

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

A

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 yield an overall positive cell potential. In this case, Co must be reduced (Eºred = +1.82V) while Na is oxidized (Eºox = -(-2.71V)).

So the total potential is:

cell = Eºred(Co) + Eºox(Na)
= 1.82 + 2.71 = 4.53 V

39
Q

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

A

The cell’s total potential is -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:

cell = Eºred(Na) + Eºox(Co)
= -2.71 + (-1.82) = -4.53 V

40
Q

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

A

ΔGº = -nFEº

Where:

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

41
Q

According to the equation

ΔGº = -nFEº

what is the sign of ΔGº for an electrolytic cell?

A

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.

42
Q

According to the equation

ΔGº = -nFEº

what is the sign of ΔGº for a galvanic cell?

A

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.

43
Q

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

A

nFEº = RT ln(Keq)

Where:

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

44
Q

According to the equation

nFEº = RT ln(Keq)

what is the value of Keq for a galvanic cell?

A

For a galvanic cell, Keq > 1.

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

45
Q

According to the equation

nFEº = RT ln(Keq)

what is the value of Keq for an electrolytic cell?

A

For an electrolytic cell, Keq < 1.

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

46
Q

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

A

nFE = -RT ln(Q/Keq)

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.

47
Q

According to the equation

nFE = -RT ln(Q/Keq)

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

A

If Q < Keq, Ecell > 0.

If Q < Keq, 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 Ecell is positive.

48
Q

According to the equation

nFE = -RT ln(Q/Keq)

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

A

If Q > Keq, Ecell < 0.

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

49
Q

According to the equation

nFE = -RT ln(Q/Keq)

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

A

If Q = Keq, Ecell = 0.

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