Physical Unit 1.11: Electrode Potentials & Cells (Redox Equilibria) Flashcards
(51 cards)
describe the +ve & -ve potential of an electrode in terms of equilibria
electrode w -ve potential:
when equlibrium lies to the left, the metal becomes -vely charged due to e-s being released & building up on the metal
electrode w +ve potential:
when equilibrium lies to the right, the metal becomes +vely charged due to e-s being used up to form metal from metal ions
what does the position of equilibrium depend on?
the metal
reactive metals tend to form Mn+ ions so -ve charge builds up on the metals
more reactive metals have -ve potentials
more unreactive metals have +ve potentials
define electrode/half-cell
a metal dipping into a solution of its ions
what electrode is used when there is no solid metal involved in the half-equation & why?
pure platinum electrode
inert
a solid metal is needed to allow the flow of e-s
what are the 3 types of electrodes?
metal electrodes
gas electrodes
redox electrodes
describe metal electrodes
metal solution surrounded by a solution of its ions
describe gas electrodes
for a gas & a solution of its ions
inert metal (usually Pt) is the electrode to allow the flow of e-s
describe redox electrodes
for 2 different ions of the same element
2 types of ion are present in solution with an inert metal electrode (usually Pt) to allow the flow of e-s
e.g. Pt(s)|Fe2+(aq), Fe3+(aq)
standard conditions: both ion solutions must be 1 mol dm-3, so [Fe2+] = 1 moldm-3 & [Fe3+] = 1 moldm-3
how do you measure the potential of an electrode?
connect half-cell to another half-cell of known potential & measure the potential difference b/w the 2 half-cells
the standard hydrogen electrode (SHE) is assigned the potential 0 & is the primary standard = the potential to which all others are compared/measured
so unknown half-cell connected to SHE with SHE on the left
what is an electrochemical cell?
2 half-cells connected
what does primary standard mean?
the potential to which all others are compared/measured
describe how 2 half-cells are joined together to give a complete circuit
2 metal electrodes are joined with a wire - to allow the transfer of e-s
the metals are dipped into a solution of their ions
the 2 solutions are joined with a salt bridge - to allow the flow of ions
a voltmeter (high resistance so low current) is often included to measure the potential difference (emf)
what can a salt bridge be made of?
a piece of filter paper soaked with a solution of unreactive ions
a tube containing unreactive ions in agar gel
e.g. KNO3 is often used in a salt bridge as K+ & NO3- are quite unreactive
(why are K+ ions unreactive?)
K+ unreactive because e- configuration is …3p6 full outer shell of e-s, which is a v stable, low-energy configuration
K+ fully oxidised
what are the standard conditions?
1.0moldm-3 solution of the ions (NB for redox electrode, 1.0moldm-3 for each ion)
298K
100kPa if half-cell includes gases
pure metal electrode
why are standard conditions required?
the position of the redox equilibrium changes when conditions change
how will the emf change if [a condition that affect position of equilibrium] is increased/decreased?
either:
emf increases/more +ve
as equilibrium shifts right so e-s used up
so more reduction happens
or
emf decreases/more -ve
as equilibrium shifts left so e-s released
so more oxidation happens
draw standard hydrogen electrode
what is the half equation?
what is the cell notation?
see booklet
2H+(aq) + 2e- <–> H2(g) convention to write as reduction
Pt(s)|H2(g)|H+(aq)
how is the conventional representation of cells written?
RO||OR
oxidation||reduction
|| indicates salt bridge
| indicates phase boundary - if no phase change, use a comma
solid metal electrode on the outside
emf =
Eθright - Eθleft
when measuring Eθ vs SHE, SHE always on the left
define secondary standard & why are they used?
= another standard electrode that has been calibrated against the SHE
SHE is difficult to use bc uses a gas, H2 is flammable & hard to keep at 100kPa
describe the redox process
metal atoms lose e-s at the -ve electrode/anode = oxidation
these e-s travel through the wire to the +ve electrode/cathode & are gained by ions to produce metal atoms = reduction
+ve electrode normally RHS
emf is not affected by the # of e-s in equation
what must be true for a reaction to be feasible?
emf must be +ve
Eθ ox. agent > Eθ red. agent
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