Electrode potentials exams!!!

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  • Created by: Hindleyc
  • Created on: 26-05-19 14:53
What does a cell have?
2 half cells
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What do the 2 half cells have to be connected with
a salt bridge
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What do Simple half cells will consist of
a metal (acts an electrode) and a solution of a compound containing that metal (eg Cu and CuSO4).
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What will These two half cells will produce
a small voltage if connected into a circuit. (i.e. become a Battery or cell).
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(Why does a voltage form) When connected together the zinc half-cell has more of a tendency to
oxidise to the Zn2+ ion and release electrons than the copper half-cell.
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What will More electrons will therefore build up on
the zinc electrode than the copper electrode.
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What is created
A potential difference is created between the two electrodes.
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What is the negative terminal and What is the positive terminal.
The zinc *****, the copper *****
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What is This potential difference measured with
a high resistance voltmeter, and is given the symbol E. The E for the above cell is E= +1.1V.
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(Why use a high resistance voltmeter?)
The voltmeter needs to be of very high resistance to stop the current from flowing in the circuit.
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So In this state what it is possible to measure
the maximum possible potential difference (E).
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What will the reactions be?
will not be occurring
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Why
because the very high resistance voltmeter stops the current from flowing.
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The salt bridge is used to
connect up the circuit.
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What conducts the charge.
The free moving ions
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What is A salt bridge is usually made from
a piece of filter paper (or material) soaked in a salt solution, usually potassium nitrate.
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What should The salt be
unreactive with the electrodes and electrode solutions.. E.g. potassium chloride would not be suitable for copper systems because chloride ions can form complexes with copper ions.
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Why is A wire is not used
because the metal wire would set up its own electrode system with the solutions.
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If the voltmeter is removed and replaced with a bulb or if the circuit is short circuited
a current flows.
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The reactions will then .
occur separately at each electrode
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The voltage will fall to
zero as the reactants are used up.
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The most positive electrode will always undergo
reduction.- Cu2+ (aq) + 2e-  Cu(s) (positive as electrons are used up)
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The most negative electrode will always undergo
oxidation.- Zn(s)  Zn2+ (aq) + 2e- (negative as electrons are given off)
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Electrochemical cells can be represented by a
cell diagram:
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EG for Zn and Cu
Zn(s) | Zn2+ (aq) | | Cu2+ (aq) | Cu (s)
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Where is Most oxidised form is put
next to the double line
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• The solid vertical line represents the
boundary between phases e.g. solid (electrode) and solution (electrolyte)
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The double line represents the
salt bridge between the two half cells
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the voltage produced is
indicated
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he more positive half cell is written on the
right if possible (but this is not essential)
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If a system does not include a metal that can act as an electrode,
then a platinum electrode must be used and included in the cell diagram
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What does It provides
a conducting surface for electron transfer.
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Why is A platinum electrode is used
because it is unreactive and can conduct electricity.
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e.g. for Fe2+ (aq)  Fe3+ (aq) + e- there is no solid conducting surface,
a Pt electrode must be used.
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The cell diagram is drawn as:
| | Fe3+ (aq), Fe2+ (aq) |Pt
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Still with more oxidised form near .A comma separates the
double line, oxidised from the reduced species.
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If the system contains several species e.g. MnO4- + 8H+ + 5e- Mn2+ + 4H2O, then in the cell diagram the balancing numbers, H+ ions and H2O can be
left out.
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eg
| |MnO4- , Mn2+ |Pt or if on left hand side Pt| Mn2+ ,MnO4- ||
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If a half equation has several physical states then
the solid vertical line should be used between each different state boundary.
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4e- +2H2O(l)+O2 (g)4OH-(aq)
||O2 |H2O,OH- |Pt
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Cl2 (g)+2e-2Cl- (aq)
||Cl2 |Cl- |Pt
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As the phase line also separates the oxidised and reduced terms
a comma is not necessary here.
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It is not possible to measure the absolute potential of a half electrode .
on its own
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It is only possible to measure the potential difference
between two electrodes.
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To measure it, it has to be connected to
another half-cell of known potential, and the potential difference between the two half-cells measured.
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by convention we can assign a relative potential to each electrode by
linking it to a reference electrode (hydrogen electrode), which is given a potential of zero Volts
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The potential of all electrodes are measured by
comparing their potential to that of the standard hydrogen electrode. The standard hydrogen electrode (SHE) is assigned the potential of 0 volts.
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The hydrogen electrode equilibrium is:
H2 (g) 2H+ (aq) + 2e-
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In a cell diagram the hydrogen electrode is represented by:
Pt |H2 (g) | H+ (aq)
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To make the electrode a standard reference electrode some conditions apply
: 1. Hydrogen gas at pressure of 100kPa 2. Solution containing the hydrogen ion at 1.0 mol dm-3 (solution is usually 1M HCl) 3. Temperature at 298K 4. Platinum electrode
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Because the equilibrium does not include a conducting metal surface,
a platinum wire is used which is coated in finely divided platinum. (The platinum black is used because it is porous and can absorb the hydrogen gas.)
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Standard conditions are needed because
the position of the redox equilibrium will change with conditions. For example, in the equilibrium: Mn+(aq) + ne- M(s) An increase in the concentration of Mn+ would move the equilibrium to the right, so making the potential more positive
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The standard hydrogen electrode is difficult to use, so often a
different standard is used which is easier to use.
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These other standards are themselves calibrated against the
SHE.
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This is known as using a
secondary standard - i.e. a standard electrode that has been calibrated against the primary standard. The common ones are: silver / silver chloride E = +0.22 V calomel electrode E = +0.27 V
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When an electrode system is connected to the hydrogen electrode system, and standard conditions apply the potential difference measured is called the
standard electrode potential,
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The standard conditions are :
•All ion solutions at 1 mol dm-3 •temperature 298K •gases at 100kPa pressure •No current flowing
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Standard electrode potentials are found in
data books and are quoted as: Li+(aq) | Li (s) E= -3.03V more oxidised form on left
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They may also be quoted as
half equations Li+ (aq) + e- Li (s) E= -3.03V but again the more oxidised form is on the left
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Note: in the electrode system containing two solutions it is necessary to use a
platinum electrode and both ion solutions must be of a 1M concentration, so [Fe2+] = 1M and [Fe3+] = 1M
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In order to calculate the Ecell, we must use
‘standard electrode potentials’ for the half cells.
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Ecell=
Erhs - Elhs (more +ve- less +ve)
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Each half cell has a
standard electrode potential value Mg2+ (aq)| Mg(s) E= -2.37V
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The most useful application of electrode potentials is
to show the direction of spontaneous change for redox reactions.
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For any two half equations The more negative half cell will always
oxidise (go backwards)
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The more positive half cell will always
reduce (go forwards)
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Mg2+ (aq) + 2e-Mg(s) E= -2.37V, Cu2+ (aq) + 2e-Cu (s) E = +0.34V. The reaction would be
Mg + Cu2+  Cu + Mg 2+
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If we want to work out the Ecell that corresponds to this spontaneous change then use
Ecell = Ered – Eox
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A spontaneous change will always have a
positive Ecell.
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To get the full equation of the reaction
add the two half reactions together, cancelling out the electrons.
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As more +ve increasing tendency for species on left to
reduce, and act as oxidising agents
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As more -ve increasing tendency for species on right
to oxidise, and act as reducing agents
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If we want to work out the Ecell from two standard electrode potentials then use
Ecell = Ered – Eox
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The most powerful reducing agents will be found at
the most negative end of the series on the right (i.e. the one with the lower oxidation number).
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The most powerful oxidising agents will be found at
the most positive end of the series on the left (i.e. the one with the higher oxidation number).
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The effects of changing conditions on E cell can be made by
applying le Chatelier’s principle.
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Ecell is a measure of
how far from equilibrium the cell reaction lies. The more positive the Ecell the more likely the reaction is to occur.
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The more positive the Ecell
the more likely the reaction is to occur.
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If current is allowed to flow
, the cell reaction will occur and the Ecell will fall to zero as the reaction proceeds and the reactant concentrations drop.
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Looking at cell reactions is a straight forward application of le Chatelier. So increasing concentration of ‘reactants’ would
increase Ecell and decreasing them would cause Ecell to decrease.
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Zn2+(aq)+2e-Zn(s) E=-0.76V Fe2+(aq)+2e-Fe(s) E=-0.44V Zn + Fe2+Fe + Zn2+ E= +0.32. IncreasingtheconcentrationofFe2+ and decreasing the concentration of Zn2+ would cause
Ecell to increase.
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Most cells are exothermic in the spontaneous direction so applying Le Chatelier to a temperature rise to these would result in a
decrease in Ecell because the equilibrium reactions would shift backwards.
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If the Ecell positive it indicates a reaction
might occur.
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There is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it does not happen. why
If the reaction has a high activation energy the reaction
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Electrochemical cells can be used as a
commercial source of electrical energy Cells can be non-rechargeable (irreversible), rechargeable and fuel cells.
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Cells are non-rechargeable when
the reactions that occur with in them are non-reversible.
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Reversible cells only work if
the product stays attached to the electrode and does not disperse
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The forward reaction occurs on
discharge giving out charge
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. Chargingcausesthereactionto
reverse
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A fuel cell uses the energy from the reaction of a
fuel with oxygen to create a voltage
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Hydrogen fuel cell (potassium hydroxide electrolyte)
4e- + 4H2O2H2 +4OH- E=-0.83V 4e- + 2H2O +O2  4OH- E=+0.4V Overall reaction 2H2 + O2  2H2O E=1.23V
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Fuel cells will maintain a
constant voltage over time because they are continuously fed with fresh O2 and H2 so maintaining constant concentration of reactants.
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This differs from ordinary cells where
the voltage drops over time as the reactant concentrations drop.
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Using standard conditions: T
he rate is too slow to produce an appreciable current. Higher temperatures are therefore used to increase rate but the reaction is exothermic so by applying le Chatelier would mean the E cell falls.
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A higher pressure can help
counteract this.
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Advantages of Fuel cells over conventional petrol or diesel-powered vehicles
(i) less pollution and less CO2. (Pure hydrogen emits only water whilst hydrogen-rich fuels produce only small amounts of air pollutants and CO2). (ii) greater efficiency
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Limitations of hydrogen fuel cells
(i) expensive (ii) storing and transporting hydrogen, in terms of safety, feasibility of a pressurised liquid and a limited life cycle of a solid ‘adsorber’ or ‘absorber’ (iii) limited lifetime (requiring regular replacement and disposal)
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and
and high prroduction costs, (iv) use of toxic chemicals in their production
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Hydrogen is readily available by the electrolysis of water,
but this is expensive. To be a green fuel the electricity needed would need to be produced from renewable resources
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Hydrogen can be stored in fuel cells
(i) as a liquid under pressure, (ii) adsorbed on the surface of a solid material, (iii) absorbed within a solid material;
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Ethanol fuel cells have also been developed. Compared to hydrogen fuel cells they have certain advantages including.
Ethanol can be made from renewable sources in a carbon neutral way. Raw materials to produce ethanol by fermentation are abundant. Ethanol is less explosive and easier to store than hydrogen.
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also
New petrol stations would not be required as ethanol is a liquid fuel.
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Equation that occurs at oxygen electrode
4e- + 4H+ +O2  2H2O E=1.23V
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Equation that occurs at ethanol electrode C2H5OH + 3H2O → 2CO2 + 12H+ + 12e-
C2H5OH + 3H2O → 2CO2 + 12H+ + 12e-
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Overall equation
C2H5OH + 3O2 → 2CO2 3h2o
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Other cards in this set

Card 2

Front

What do the 2 half cells have to be connected with

Back

a salt bridge

Card 3

Front

What do Simple half cells will consist of

Back

Preview of the front of card 3

Card 4

Front

What will These two half cells will produce

Back

Preview of the front of card 4

Card 5

Front

(Why does a voltage form) When connected together the zinc half-cell has more of a tendency to

Back

Preview of the front of card 5
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