14.1

Question 1

what is the standard electrode potential

Answer 1

measuring how easily a substance loses electrons

Question 2

what is the absolute potential difference

Answer 2

potential difference between the metal and the solution

Question 3

why is it not possible to measure directly the absolute potential difference and how can it be solved

Answer 3

because in order to create a circuit there must be another metal dipped in the solution which will interfere with the results
instead a reference electrode is used to measure the potential difference between the metal electrode and the reference electrode

Question 4

what is the reference electrode

Answer 4

standard hydrogen electrode

Question 5

describe the standard hydrogen electrode

Answer 5

hydrogen gas at 100kPa bubbled over platinum foil dipped in 1mol/cm^3 HCl solution at 298k

Question 6

why is the platinum foil for the standard hydrogen electrode covered in porous platinum

Answer 6

it has a large surface area and allows equilibrium between hydrogen ions in solution and hydrogen gas to be reached quickly

Question 7

what are the standard conditions for electrodes

Answer 7

100kPa
298k
1mol/dm^3 solution

Question 8

what is a salt bridge needed for in a complete cell

Answer 8

completes the electrical circuit by allowing the movement of ions

Question 9

what creates a salt bridge

Answer 9

paper soaked in a concentrated solution of potassium nitrate (or any ionic salt which will not interfere with the components of the half cell)

Question 10

how do electrons move in positive electrodes

Answer 10

electrons flow from the hydrogen electrode to the metal electrode

Question 11

how do electrons move in a negative electrode

Answer 11

electrons flow from the metal electrode to the hydrogen electrode

Question 12

where does the position of equilibrium lie when the E^o value is more negative and is the species a more powerful reducing or oxidizing agent

Answer 12

to the left
reducing agent

Question 13

where does the position of equilibrium lie when the E^o value is more positive and is the species a more powerful reducing or oxidizing agent

Answer 13

to the right
oxidizing agent

Question 14

what is the electromotive force (emf)

Answer 14

the standard electrode potential of a half cell measured under standard conditions of 298K, 100kPa, 1mol/dm^3 connected to a standard hydrogen electrode

Question 15

define standard electrode potential

Answer 15

the standard electrode potential of a half cell is the emf of a cell containing the half-cell connected to the standard hydrogen electrode under standard states

Question 16

what is the cell diagram for the zinc and copper cells

Answer 16

Zn(s)/Zn^2+(aq) :: Cu^2+(aq)/ Cu(s)

Question 17

what does :: represent in a cell diagram

Answer 17

the salt bridge

Question 18

how do electrons flow between half cells

Answer 18

from the more negative E^o value half cell to the less negative E^o value half cell

Question 19

is this reaction thermodynamically feasible: Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)
Zn^2+(aq) + 2e- <=> Zn(s) E^o = -0.76 V
Cu^2+(aq) + 2e- <=> Cu(s) E^o = +0.34 V

Answer 19

yes because Zn has a more negative E^o value therefor equilibrium lies to the left and is releasing electrons which are accepted by Cu as it has a more positive E^o value therefor equilibrium lies to the right

Question 20

is this reaction thermodynamically feasible: Zn(s) + 2H+(aq) -> Zn^2+(aq)
Zn^2+(aq) + 2e- <=> Zn(s) E^o = -0.76 V
2H+(aq) + 2e- <=> H2(g) E^o = 0.00 V

Answer 20

yes because Zn has a more negative E^o value therefor equilibrium lies to the left and is releasing electrons which are accepted by H as it has a more positive E^o value therefor equilibrium lies to the right

Question 21

is this reaction thermodynamically feasible: Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)
Cu^2+(aq) + 2e- <=> Cu(s) E^o = +0.34 V
2H+(aq) + 2e- <=> H2(g) E^o = 0.00 V

Answer 21

no because Cu has a more positive E^o value than H so it is not thermodynamically feasible as the two equilibria move in opposite directions to those required for the reaction

Question 22

is this reaction thermodynamically feasible: Cu^2+(aq) + H2(g) -> Cu(s) + 2H+(aq)
Cu^2+(aq) + 2e- <=> Cu(s) E^o = +0.34 V
2H+(aq) + 2e- <=> H2(g) E^o = 0.00 V

Answer 22

yes because H has a more negative E^o value therefor equilibrium lies to the left and is releasing electrons which are accepted by Cu as it has a more positive E^o value therefor equilibrium lies to the right

Question 23

is this reaction thermodynamically feasible:
MnO2(s) + 4HCl(aq) -> Mn^2+(aq) + 2Cl-(aq) + 2H2O(l) + Cl2(g)
MnO2(s) + 4H+(aq) + 2e- <=> Mn^2+(aq) + 2H2O(l) E^o = +1.23
Cl2(g) + 2e- <=> 2Cl-(aq) E^o = +1.36

Answer 23

no because Cl has a more positive E^o cell value than MnO2 so it is not thermodynamically feasible as the two equilibria move in opposite directions to those required for the reaction

Question 24

How can you make this reaction thermodynamically feasible
MnO2(s) + 4HCl(aq) -> Mn^2+(aq) + 2Cl-(aq) + 2H2O(l) + Cl2(g)
MnO2(s) + 4H+(aq) + 2e- <=> Mn^2+(aq) + 2H2O(l) E^o = +1.23
Cl2(g) + 2e- <=> 2Cl-(aq) E^o = +1.36

Answer 24

increasing the concentration of HCl will shift the position of equilibrium to the left and the E^o value becomes less positive and the reaction becomes thermodynamically feasible (not under standard conditions)

Question 25

why may a reaction not occur even if the standard electrode potentials indicate the reaction is thermodynamically feasible

Answer 25

the activation energy may be too large
the reaction may not be taking place under standard conditions

Question 26

explain the disproportionation of copper using electrode potentials

Answer 26

Cu^2+(aq) + e- <=> Cu+(aq) E^o = +0.15V
Cu+(aq) + e- <=> Cu(s) E^o = +0.52V
the first equilibrium is more negative therefor it goes to the left (oxidation) and the second equilibrium is more positive therefor it goes to the right (reduction)

Question 27

what is the equation relating #G and E^ocell

Answer 27

G = -nFE^ocell

n - number of moles involved in reaction
F - faraday constant

Question 28

what is the equation relating lnK and E^ocell

Answer 28

lnK = nFE^ocell/RT