Electrochemistry - Physical Chemistry

Front 1

A ____ cell is the experimental apparatus for generating ____ through the use of a spontaneous ____ reaction.

Back 1

A galvanic/voltaic cell is the experimental apparatus for generating electricity through the use of a spontaneous redox reaction.

Front 2

An electric current flows from the ____ to the ____ for spontaneous reactions because there is a difference in electrical ____ between the electrodes.

Back 2

An electric current flows from the anode to the cathode for spontaneous reactions because there is a difference in electrical potential between the electrodes.

Front 3

The voltage of a cell depends not only on the nature of electrodes and the ions, but also on the ____ of the ions and the ____ at which the cell is operated.

Back 3

The voltage of a cell depends not only on the nature of electrodes and the ions, but also on the concentrations of the ions and the temperature at which the cell is operated.

Front 4

The measured ____ is the maximum voltage that a cell can achieve and is used to calculate the maximum amount of ____ ____ that can be obtained from the chemical reaction.

Back 4

The measured EMF (electromotive force) is the maximum voltage that a cell can achieve and is used to calculate the maximum amount of electrical energy that can be obtained from the chemical reaction.

Front 5

State an expression for the standard Gibbs free energy change, ΔG^∅, in terms of the standard cell potential, E^∅[cell].

Back 5

Where:

n = stoichiometric no. of electrons transferred in redox reaction

F = Faraday's constant (96485 C mol^-1)

Front 6

State an expression for the standard Gibbs free energy change, ΔG^∅, in terms of the redox reaction equilibrium constant, K.

Back 6

Where:

R = ideal gas constant (8.3145 J K^-1 mol^-1)

T = temperature / K

Front 7

Predict the spontaneous cell reaction for these two half cells at 298 K.

Back 7

Front 8

Calculate the equilibrium constant for this reaction at 298 K.

Back 8

6.5 x 10^9

Front 9

Calculate the standard Gibbs free energy change for this reaction at 298 K.

Back 9

2.53 x 10^3 kJ mol^-1

Front 10

The standard Gibbs free energy change for this reaction is -206 kJ mol^-1. Calculate the standard cell potential.

Back 10

0.71 V

Front 11

Calculate ΔG^∅ for this reaction at 298 K.

Back 11

-411 kJ mol^-1

Front 12

Calculate the equilibrium constant for this reaction at 298 K.

Back 12

1.14 x 10^-42

Front 13

The cathode is where ____ occurs and the anode is where ____ occurs.

Back 13

The cathode is where reduction occurs and the anode is where oxidation occurs.

Front 14

____ difference between the ____ of a cell is the measure of the tendency of the cell reaction to take place.

Back 14

Potential difference between the electrodes of a cell is the measure of the tendency of the cell reaction to take place.

Front 15

The more ____ the cell potential, the greater the tendency for the reaction to proceed to the right.

Back 15

The more positive the cell potential, the greater the tendency for the reaction to proceed to the right.

Front 16

____ ____ can differ significantly from a value of 1, due to the strong ____ interactions in electrolyte solutions, resulting in marked deviations from ____ even in ____ systems.

Back 16

Activity coefficients can differ significantly from a value of 1, due to the strong ionic interactions in electrolyte solutions, resulting in marked deviations from ideality even in dilute systems.

Front 17

Standard cell potentials refer to cells in which all dissolved substances are at ____ ____ (i.e. ____ mol kg^-1), and any gases are at ____ ____ pressure.

Back 17

Standard cell potentials refer to cells in which all dissolved substances are at unit activity (i.e. 1 mol kg^-1), and any gases are at 1 atm pressure.

Front 18

State the Nernst equation relating cell potential to standard cell potential for this reaction.

Back 18

Where:

E = cell potential / V

E^∅ = standard cell potential / V

R = ideal gas constant (8.3145 J K^-1 mol^-1)

T = temperature / K

n = stoichiometric no. of electrons transferred

F = Faraday constant (96485 C mol^-1)

a = activity

Front 19

The Nernst equation is only valid for ____ ____ solutions as ions of ____ charge tend to associate in to ____ ____ in ____ solutions, reducing the number of ions free to donate/accept electrons at an ____. Therefore, the Nernst equation cannot accurately predict half-cell potentials for solutions whose concentration is greater than ____ M.

Back 19

The Nernst equation is only valid for dilute ionic solutions as ions of opposite charge tend to associate in to ion pairs in concentrated solutions, reducing the number of ions free to donate/accept electrons at an electrode. Therefore, the Nernst equation cannot accurately predict half-cell potentials for solutions whose concentration is greater than 0.001 M.

Front 20

In an electrochemical cell a zinc electrode is immersed in a solution containing Zn^2+ ions (a = 0.1 mol kg^-1) and separated by a porous membrane, a copper electrode is immersed in a solution of Cu^2+ ions (a = 0.01 mol kg^-1). Determine the spontaneous reaction equation.

Back 20

Front 21

Predict whether this reaction would proceed spontaneously as written as 298 K. (Assume activity coefficients of 1)

Back 21

The reaction is not spontaneous.

Front 22

Determine the spontaneous reaction equation for this electrochemical cell at 298 K.

Back 22

Front 23

Determine the cell potential for this Daniel cell at 313 K.

Back 23

0.80 V

Front 24

Define a concentration cell.

Back 24

A concentration cell is a limited form of a galvanic cell that has two equivalent half-cells of the same material differing only in concentrations.

Front 25

A concentration cell produces a small ____ as it attempts to reach ____. This occurs when the concentrations of reactant in both cells are ____.

Back 25

A concentration cell produces a small voltage as it attempts to reach equilibrium. This occurs when the concentrations of reactant in both cells are equal.

Front 26

State the Nernst equation for a concentration cell.

Back 26

Where:

E = cell potential / V

R = ideal gas constant (8.3145 J K^-1 mol^-1)

T = temperature / K

n = stoichiometric no. of electrons transferred

F = Faraday constant (96485 C mol^-1)

c = concentration

Front 27

Calculate the cell potential for this cell at 298 K.

Back 27

0.03 V

Front 28

State three applications of concentrations cells.

Back 28

Any three from:

i) Nerve signalling

ii) Ion pumps across cell membranes

iii) Energy production and storage in cells

iv) Measuring solubility products

v) pH meters

vi) Ion selective electrodes

vii) Corrosion

Front 29

____ is a natural phenomenon which attacks a metal by chemical or electrochemical action and converts it in to a metallic compound such as an ____, ____ or ____.

Back 29

Corrosion is a natural phenomenon which attacks a metal by chemical or electrochemical action and converts it in to a metallic compound such as an oxide, hydroxide or sulphate.

Front 30

Metals corrode because we use them in environments in which they are ____ ____. Only ____ and the ____ metals are found in their metallic state in nature. All other metals we use are processed from ____ or ____ in to metal. Some metals form protective ____ on their surface to prevent or slow down corrosion.

Back 30

Metals corrode because we use them in environments in which they are chemically unstable. Only copper and the precious metals are found in their metallic state in nature. All other metals we use are processed from minerals or ores in to metal. Some metals form protective films on their surface to prevent or slow down corrosion.

Front 31

____ ____ ____ occurs when two or more areas of a metallic surface are in contact with different concentrations of the same solution.

Back 31

Concentration cell corrosion occurs when two or more areas of a metallic surface are in contact with different concentrations of the same solution.

Front 32

Describe metal ion concentration cell corrosion.

Back 32

In the presence of water, a high concentration of metal ions will exist under faying surfaces and a low concentration of metal ions will exist adjacent to the crevice created by the faying surfaces. An electrical potential will exist between the two points. The area of the metal in contact with the high concentration of metal ions will be cathodic and will be protected, whereas the area of the metal in contact with the low metal ion concentration will be anodic and corroded.

Front 33

Describe oxygen concentration cell corrosion.

Back 33

Water in contact with a metal surface usually contains dissolved oxygen. An oxygen cell can develop at any point where the oxygen in the air is not allowed to diffuse uniformly in to the solution, thereby creating a difference in oxygen concentration between two points. Corrosion will occur at the area of low-oxygen concentration which is anodic.

Front 34

Describe active-passive concentration cell corrosion.

Back 34

On metals that depend on a tightly adhering passive film (usually an oxide) for corrosion protection, salt that deposits on the metal surface in the presence of water, in areas where the passive film is broken, will cause the active metal beneath the film to be exposed to corrosive attack. An electrical potential will develop between the large area of the cathode (passive film) and the small area of the anode (active metal). Rapid pitting of the active metal will result.

Front 35

Calculate the cell potential for this cell at 298 K.

Back 35

0.12 V

Front 36

In a beaker a copper electrode is dipped in a solution of Cu^2+ ions with a concentration of 0.1 M the electrode is connected to another copper electrode dipped in to a solution of Cu^2+ of unknown concentration. The two beakers are linked by a salt bridge and the measured potential of the cell is 0.75 V at 298 K. What is the unknown concentration of Cu^2+? Assume the unknown is anodic.

Back 36

4.29 x 10^-27 M

Front 37

State an expression for pH.

Back 37

Where:

a = activity / mol dm^-3

Front 38

If the concentration of active hydrogen and active hydroxyl ions in a solution is of the same quantity, the solution is ____.

Back 38

If the concentration of active hydrogen and active hydroxyl ions in a solution is of the same quantity, the solution is neutral.

Front 39

If a solution has a higher concentration of active hydrogen ions than that of hydroxyl ions, the solution is ____.

Back 39

If a solution has a higher concentration of active hydrogen ions than that of hydroxyl ions, the solution is acidic.

Front 40

If a solution has a higher concentration of active hydroxyl ions than that of hydrogen ions, the solution is ____.

Back 40

If a solution has a higher concentration of active hydroxyl ions than that of hydrogen ions, the solution is basic.

Front 41

Describe the visual method of pH determination.

Back 41

A colour comparison with pH-sensitive indicator paper (litmus) to a standard colour scale.

Front 42

Describe the photometric method of pH determination.

Back 42

Use of a spectrophotometer to measure the wavelength of a coloured pH-sensitive solution.

Front 43

Describe the potentiometric method of pH determination.

Back 43

An electrochemical measurement, measuring the EMF (electromotive force) created by a chemical reaction, such as that which takes place between metals and dissolved salts.

Front 44

An increase of 1 unit in pH decreases the measured potential of a solution by ____ mV.

Back 44

An increase of 1 unit in pH decreases the measured potential of a solution by 59.2 mV.

Front 45

State the three components of a pH measurement system.

Back 45

i) The measuring electrode (a pH-sensitive electrode)

ii) The reference electrode

iii) A high-impedance voltmeter

Front 46

Fill in the labels on this glass electrode.

Back 46

Front 47

The glass shaft on a glass electrode is highly resistant to ____, ____ solutions.

Back 47

The glass shaft on a glass electrode is highly resistant to hot, alkaline solutions.

Front 48

The pH-sensitive part of a glass electrode is the ____-shaped electrode tip, the ____ ____.

Back 48

The pH-sensitive part of a glass electrode is the cylindrically-shaped electrode tip, the glass membrane.

Front 49

pH measurement cannot be more accurate than the ____ used, typically ±____ pH units.

Back 49

pH measurement cannot be more accurate than the standards used, typically ±0.01 pH units.

Front 50

State and explain the five sources of error in pH measurement using a glass electrode.

Back 50

i) Equilibration time - the time it takes for the electrode to equilibrate with the solution can range from 30 seconds to minute depending on conditions e.g. adequate stirring.

ii) Hydration of glass - a dry electrode requires several hours of soaking before it responds correctly to hydrogen ions.

iii) Temperature - the pH meter should be calibrated at the same temperature at which the measurement will be made.

iv) Junction potential - if the analyte solution is different from the buffer inside the pH electrode, the junction potential will change even if the pH of the two solutions is the same.

v) Junction potential drift - caused by silver precipitating in the plug - can be compensated for by recalibration

Front 51

The glass electrode used to measure pH is the most common example of an ____-____ ____. Ideally, it should respond only to the intended ion and be ____ by other species - in practice there is always ____.

Back 51

The glass electrode used to measure pH is the most common example of an ion-selective electrode. Ideally, it should respond only to the intended ion and be unaffected by other species - in practice there is always interference.

Front 52

Ion-selective electrodes are fundamentally different from metal electrodes as they don't depend on ____ processes. An ideal ISE is a ____ ____ across which only the ion can ____. The ____ ____ difference that results from the diffusion of the ion is related to the activity of the ion on the two sides of the ____.

Back 52

Ion-selective electrodes are fundamentally different from metal electrodes as they don't depend on redox processes. An ideal ISE is a thin membrane across which only the ion can migrate. The equilibrium potential difference that results from the diffusion of the ion is related to the activity of the ion on the two sides of the membrane.

Front 53

The current in a galvanic cell flows in the direction of a ____ reaction. We must apply an ____ ____ in the opposite direction of this current flow to make the opposite reaction occur. This process is called ____.

Back 53

The current in a galvanic cell flows in the direction of a spontaneous reaction. We must apply an external current in the opposite direction of this current flow to make the opposite reaction occur. This process is called electrolysis.

Front 54

State the three common causes of over-potential.

Back 54

i) Resistance in the electrical circuit

ii) The rate of electron transfer is limited by electron transfer at the electrode interface

iii) If a half-reaction has a high barrier for electron transfer then the rate is slowed.

Front 55

Cations in solution that have a more positive ____ ____ than water can be reduced. Anions in solution that have a more positive ____ ____ than water can be oxidised.

Back 55

Cations in solution that have a more positive reduction potential than water can be reduced. Anions in solution that have a more positive oxidation potential than water can be oxidised.

Front 56

State Faraday's first law of electrolysis.

Back 56

The mass of a substance liberated at an electrode during electrolysis is proportional to the quantity of charge (in Coulombs) passing through the electrolyte.

Front 57

State Faraday's second law of electrolysis.

Back 57

The number of Coulombs needed to liberate one mole of different products are in whole number ratios.

Front 58

State an expression relating charge to current.

Back 58

Where:

Q = charge / C

I = current / A

t = time / s

Front 59

What does the Faraday constant describe?

Back 59

The experimental charge on one mole of electrons.

Front 60

In the electro-refining of copper, what mass of copper(II) is deposited in one hour by a current of 1.62 A?

Back 60

1.92 g

Front 61

In the electrolysis of nickel, what mass of nickel is deposited in 2.5 hours by a current of 0.84 A?

Back 61

2.30 g

Front 62

Predict the spontaneous cell reaction for the combination of these two half-cells and calculate the standard cell potential.

Back 62

2.74 V

Front 63

Calculate the equilibrium constant for this reaction at 298 K.

Back 63

5.63 x 10^51

Front 64

Calculate the standard Gibbs free energy change for this reaction at 298 K.

Back 64

360 kJ mol^-1 (make sure this answer is positive!)

Front 65

In an electrochemical cell, an iron electrode is immersed in a solution containing iron(II) ions (a = 0.1 mol kg^-1) and, separated by a porous membrane, a copper electrode is immersed in a solution of copper(II) ions (a = 0.01 mol kg^-1). Determine the spontaneous cell reaction at 298 K.

Back 65

Front 66

Predict whether this reaction would proceed spontaneously as written at 298 K. Assume an activity coefficient of 1.

Back 66

Yes, the reaction would proceed spontaneously as written.

Front 67

In a beaker a copper electrode (the cathode) is dipped in to a solution of copper(II) ions with a concentration of 0.1 M. The electrode is connected to another copper electrode (the anode) dipped in to a solution of copper(II) ions with a concentration of 0.02 M. The two beakers are linked by a salt bridge. What is the cell potential at 298 K? Assume an activity coefficient of 1.

Back 67

0.021 V

Front 68

In the electro-refining of silver, what mass of silver(I) is deposited in 3.5 hours by a current of 3.8 A?

Back 68

53.5 g

Front 69

Solubility products are ____ ____ for solubility.

Back 69

Solubility products are equilibrium constants for solubility.

Front 70

State an expression for the solubility product, K[sp], of barium sulphate.

Back 70

Front 71

Why is the concentration of the solid left out of the solubility product expression?

Back 71

Solubility products describe heterogeneous equilibria (two separate solid and aqueous phases)

Front 72

State an expression for the solubility product, K[sp], of calcium phosphate.

Back 72

Front 73

The solubility of barium sulphate at 298 K is 1.05 x 10^-5 M. Calculate the solubility product.

Back 73

1.10 x 10^-10 mol^2 dm^-6

Front 74

The solubility of magnesium hydroxide at 298 K is 1.71 x 10^-4 M. Calculate the solubility product.

Back 74

2.00 x 10^-11 mol^3 dm^-9

Front 75

If the solubility product of calcium phosphate is 2.07 x 10^-33 mol^5 dm^-15 at 298 K, calculate its solubility in mol dm^-3 at that temperature.

Back 75

2.03 x 10^-7 mol dm^-3

Front 76

Evaluate the solubility of silver chloride at 298 K from cell potential data.

Back 76

1.25 x 10^-5 mol kg^-1

Front 77

State an expression for the solubility product, K[sp], and its units for silver iodide.

Back 77

mol^2 dm^-6

Front 78

State an expression for the solubility product, K[sp], and its units for mercury(I) sulphide.

Back 78

mol^3 dm^-9

Front 79

State an expression for the solubility product, K[sp], and its units for iron(III) hydroxide.

Back 79

mol^4 dm^-12

Front 80

State an expression for the solubility product, K[sp], and its units for silver chromate.

Back 80

mol^3 dm^-9

Front 81

Calculate the molar solubility of silver carbonate in water at 298 K, given a solubility product of 6.2 x 10^-12 mol^3 dm^-9 at this temperature.

Back 81

1.46 x 10^-4 mol dm^-3

Front 82

Calculate the molar solubility of iron(III) hydroxide in water at 298 K, given a solubility product of 2.0 x 10^-39 mol^4 dm^-12 at this temperature.

Back 82

1.61 x 10^-10 mol dm^-3

Front 83

What effect would the addition of sodium chloride have on the position of this equilibrium? Name this effect.

Back 83

Shift of position to the left (to produce more solid)

This is the common ion effect.

Front 84

Why isn't the concentration of chloride ions produced by dissolution of lead chloride taken in to account when calculating the total concentration of chloride ions when it is added to a solution of sodium chloride?

Back 84

The amount of chloride ions produced by dissolution of the lead chloride is infinitesimal compared to the concentration already present from the sodium chloride.

Front 85

Calculate the molar solubility of lead chloride in 0.1 M sodium chloride at 298 K, given a solubility product of 1.7 x 10^-5 mol^3 dm^-9 at this temperature.

Back 85

1.7 x 10^-3 mol dm^-3

Front 86

Describe and explain the non-common ion effect.

Back 86

At low concentrations of a soluble salt added to a solution of a sparingly soluble salt, the solubility of the latter is increased. This is due to the addition of a greater oppositely charged ionic atmosphere surrounding each ion of the sparingly soluble salt. This slightly lowers the energy of these ions allowing the salt to dissolve more.

Front 87

Calculate the molar lead(II) ion concentration of a solution of lead chloride in 0.5 M sodium chloride at 298 K, given a solubility product of 1.7 x 10^-5 mol^3 dm^-9 at this temperature.

Back 87

6.8 x 10^-5 mol dm^-3

Front 88

Calculate the molar lead(II) ion concentration of a solution of lead chloride in 1 M sodium chloride at 298 K, given a solubility product of 1.7 x 10^-5 mol^3 dm^-9 at this temperature.

Back 88

1.7 x 10^-5 mol dm^-3

Front 89

Predict the standard cell potential for the combination of these half-cells. State the spontaneous cell equation.

Back 89

E^∅[cell] = 0.14 V

Front 90

Determine ΔG for this reaction at 298 K.

Back 90

120 kJ mol^-1

Front 91

In an electrochemical cell, a copper electrode is immersed in a solution containing copper(II) ions (a = 1 mol kg^-1) and separated by a porous membrane a barium electrode is immersed in a solution containing barium(II) ions (a = 0.01 mol kg^-1). Determine the cell potential.

Back 91

3.3 V

Front 92

Determine the equilibrium constant for this reaction at 298 K. Write down the half-reactions and denote which half-cell is oxidised and which is reduced.

Back 92

K = 9.77 x 10^-26

Front 93

The solubility of barium sulphate at 298 K is 1.05 x 10^-5 mol dm^-3. Calculate the solubility product.

Back 93

1.10 x 10^-10 mol^2 dm^-6

Front 94

Suppose you tried to dissolved some lead(II) chloride in 0.1 M sodium chloride as well as in pure water. Calculate the factor by which the concentration of lead(II) ions differs between the two solutions.

Back 94

The concentration has fallen by a factor of 9.53.