14.1 Flashcards
what is the standard electrode potential
measuring how easily a substance loses electrons
what is the absolute potential difference
potential difference between the metal and the solution
why is it not possible to measure directly the absolute potential difference and how can it be solved
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
what is the reference electrode
standard hydrogen electrode
describe the standard hydrogen electrode
hydrogen gas at 100kPa bubbled over platinum foil dipped in 1mol/cm^3 HCl solution at 298k
why is the platinum foil for the standard hydrogen electrode covered in porous platinum
it has a large surface area and allows equilibrium between hydrogen ions in solution and hydrogen gas to be reached quickly
what are the standard conditions for electrodes
100kPa
298k
1mol/dm^3 solution
what is a salt bridge needed for in a complete cell
completes the electrical circuit by allowing the movement of ions
what creates a salt bridge
paper soaked in a concentrated solution of potassium nitrate (or any ionic salt which will not interfere with the components of the half cell)
how do electrons move in positive electrodes
electrons flow from the hydrogen electrode to the metal electrode
how do electrons move in a negative electrode
electrons flow from the metal electrode to the hydrogen electrode
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
to the left
reducing agent
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
to the right
oxidizing agent
what is the electromotive force (emf)
the standard electrode potential of a half cell measured under standard conditions of 298K, 100kPa, 1mol/dm^3 connected to a standard hydrogen electrode
define standard electrode potential
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
what is the cell diagram for the zinc and copper cells
Zn(s)/Zn^2+(aq) :: Cu^2+(aq)/ Cu(s)
what does :: represent in a cell diagram
the salt bridge
how do electrons flow between half cells
from the more negative E^o value half cell to the less negative E^o value half cell
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
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
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
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
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
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
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
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
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
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
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
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)
