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Chemistry > 14.1 > Flashcards

14.1 Flashcards

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1
Q

what is the standard electrode potential

A

measuring how easily a substance loses electrons

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2
Q

what is the absolute potential difference

A

potential difference between the metal and the solution

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3
Q

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

A

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

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4
Q

what is the reference electrode

A

standard hydrogen electrode

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5
Q

describe the standard hydrogen electrode

A

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

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6
Q

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

A

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

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7
Q

what are the standard conditions for electrodes

A

100kPa
298k
1mol/dm^3 solution

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8
Q

what is a salt bridge needed for in a complete cell

A

completes the electrical circuit by allowing the movement of ions

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9
Q

what creates a salt bridge

A

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

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10
Q

how do electrons move in positive electrodes

A

electrons flow from the hydrogen electrode to the metal electrode

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11
Q

how do electrons move in a negative electrode

A

electrons flow from the metal electrode to the hydrogen electrode

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12
Q

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

A

to the left
reducing agent

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13
Q

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

A

to the right
oxidizing agent

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14
Q

what is the electromotive force (emf)

A

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

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15
Q

define standard electrode potential

A

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

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16
Q

what is the cell diagram for the zinc and copper cells

A

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

17
Q

what does :: represent in a cell diagram

A

the salt bridge

18
Q

how do electrons flow between half cells

A

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

19
Q

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

A

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

20
Q

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

A

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

21
Q

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

A

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

22
Q

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

A

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

23
Q

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

A

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

24
Q

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

A

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)

25
Q

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

A

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

26
Q

explain the disproportionation of copper using electrode potentials

A

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)

27
Q

what is the equation relating #G and E^ocell

A

G = -nFE^ocell

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

28
Q

what is the equation relating lnK and E^ocell

A

lnK = nFE^ocell/RT

Chemistry (46 decks)

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