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Chemistry > 20. Electrochemistry > Flashcards

20. Electrochemistry Flashcards

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1
Q

What is electrolysis?

A

The decomposition of a compound into its elements by an electric current. Carried out in an electrolysis cell.

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2
Q

What can be produced from electrolysis?

A

Reactive metals can be extracted from their ores (too high in reactivity series to be extracted by heating with carbon). Metals formed are either deposited as a layer on the cathode or form a molten layer in the cell.
Non-metals like chlorine can be produced. Gases are bubbled off.
Metals can be purified.

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3
Q

What is an electrolyte?

A

The compound that is to be decomposed. It can be a molten ionic compound or a conc. aqueous solution of ions.

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4
Q

What is an electrode?

A

Rods made from either graphite or a metal, used to conduct electricity to and from the electrolyte.
Anode = positive, cathode = negative.

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5
Q

What is the power supply in an electrolysis cell?

A

Direct current.

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6
Q

The mass deposited at an electrode is proportional to…

A

Time over which an electric current passes, and the strength of the current.

AKA the quantity of electricity passing through the electrolyte, given by Q = It.

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7
Q

What is a Faraday?

A

The quantity of electric charge carried by 1 mole of electrons or singly charged ions - 96500C/mol.

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8
Q

How is the Avogadro constant related to the Faraday?

A

F = Le

  • F = charge on 1 mole of electrons
  • e = charge on 1 electron (1.6 x 10^-19)
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9
Q

What is the electrolytic method for finding the charge on an electron?

A

Electrolysis cell with Cu cathode and anode. Ammeter, d.c. power supply and variable resistor (maintains a constant current) connected in series.

  • Weigh the electrodes separately.
  • Apply a constant electric current for a measured time period.
  • Wash and dry the electrodes (distilled water, then propanone) and re-weigh.
  • The cathode should have increased in mass, the anode should have decreased - measure the loss in mass.
  • Calculation should support F = Le.
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10
Q

When does a redox EQM exist?

A

Between chemically related species in different oxidation states, eg. copper rod in solution of its ions. It is established when the rate of electron gain = rate of electron loss.

There are two opposing reactions:

  • metal atoms from the rod entering the solution as metal ions, leaving electrons behind on the surface of the rod
  • copper ions in solution accepting electrons from the rod, deposited on the surface of the rod as copper metal.
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11
Q

What does reactivity have to do with redox EQM?

A

For unreactive metals, EQM lies further to the right (compared with other metal EQM). For reactive metals, it lies further to the left (eg. V 2+ ions).

The more reactive, the harder it is to reduce, so the further the EQM will lie to the left.

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12
Q

What is electrode/reduction potential?

A

When a metal is placed into a solution of its ions, a voltage (electric potential) is set up between the two. This cannot be measured directly but the difference between this and another system can be (uses the standard hydrogen electrode).

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13
Q

How is this electric potential caused?

A

Through the formation of an electrical double layer when an element is placed in a solution of its ions, eg. Zn and Zn2+.

  • some atoms on Zn metal’s surface release electrons and are converted to Zn 2+, which go into solution.
  • excess electrons are left on Zn’s surface - these attract the excess Zn 2+ ions that are in the surrounding solution.
  • a double layer is formed - this charge buildup causes a voltage between the metal atoms and ions in solution.
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14
Q

Describe the standard hydrogen electrode.

A
  • H+ ions in solution (1.00 mol/dm3)
  • H2 gas at 1 ATM pressure (forms EQM with H+)
  • inert platinum electrode covered with platinum black (finely divided Pt) in contact with the EQM.

The finely divided Pt allows for H2 and H+ to be in contact and set up an EQM quickly.
Standard electrode potentials for all half-cells are measured against this - its own EΘ is 0.00V.

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15
Q

What does an electrode potential tell us?

A

How easy it is to reduce a substance. It refers to the reduction reaction, so electrons always appear on the LEFT. More positive = easier to reduce species on the left. Metal on right is relatively unreactive / poor reducing agent (for the other half-cell).

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16
Q

What is standard electrode potential?

A

For a half-cell; the voltage measured under standard conditions with the standard hydrogen electrode as the other half-cell.
It is affected by temperature, pressure and concentration so STD conditions must be used.

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17
Q

Describe a metal/metal ion half-cell.

A

Metal in contact with a solution of its ions, eg. Cu2+/Cu, EΘ +0.34V.
- It is the positive pole - Cu2+ is easier to reduce than H+ (more likely to accept electrons, so it will gain electrons from H+/H2 half-cell and the H+/H2 half-cell will lose electrons to the Cu2+/Cu half-cell).

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18
Q

Describe a nonmetal/nonmetal ion half-cell.

A

Nonmetal in contact with a solution of its ions, eg. Cl2/Cl-, EΘ +1.36V.

  • It is the positive pole; H+/H2 is the negative pole.
  • uses Pt wire/foil as electrode, surface of which EQM is established. It is in contact with the element and the aqueous solution of its ions.
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19
Q

Describe an ion/ion half-cell.

A

They have ions of the same element with different oxidation states, eg. Fe3+/Fe2+, EΘ +0/77V.
- Pt or graphite electrode.

eg2. MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O.
- H+ needed for the reduction reaction so all three ions are present in solution (1mol/dm3 each).

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20
Q

What is the standard cell potential?

A

The difference in EΘ values between two half-cells in an electrochemical cell.
The less positive is always subtracted from the more positive.

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21
Q

How can you predict the direction of electron flow?

A

eg. Zn2+/Zn (EΘ -0.76V), Ag+/Ag (EΘ +0.80V).
- Zn metal will lose electrons to the Ag+/Ag half-cell - these will travel through wires in the external circuit to this half-cell to reduce Ag+.

Electrons will flow from the negative pole to the positive pole.

22
Q

How can you predict the occurrence of a reaction?

A

In half-equations, the more oxidised species is on the left (more easily reduced), and vice versa.
EΘ can predict whether the reaction will proceed in its forward (reduction) direction or its reverse (oxidation) reaction.

23
Q

What is feasibility?

A

The feasibility is how likely a reaction is to occur. According to the ionic equation:
Zn + Cu2+ -> Zn2+ + Cu, we can see that placing Zn metal in 1M Cu2+ solution is feasible, but doing the reverse is not.
Using the reduction potential list, we can see the direction of reaction can be given by a clockwise pattern (reactant, product, reactant, product).

24
Q

How do you calculate the occurrence of a reaction?

A

1) Write the half-equations.
2) Identify the stronger oxidising and reducing agents.
3) The stronger O agent will react with the stronger R agent in a feasible reaction.
4) Combine the two half-equations.

25
Q

How do you find the relative voltage without knowing the better O/R agents?

A

1) Write the proposed reaction.
2) Write the corresponding half-equations.
3) Find EΘ for each reaction, flip the sign for the reaction that is reversed to oxidation.
4) Add the two voltages (we are combining the half-equations).
5) If the voltage is positive then the reaction is feasible.

26
Q

As EΘ values for reduction reactions get less positive…

A

The species on the left become weaker oxidising agents (accept electrons less readily) and the species on the right become stronger reducing agents (release electrons more readily).

27
Q

How do you reduce/oxidise a substance with a very low/high EΘ value?

A

eg. Zn2+ must be reacted with a stronger R agent, eg. Mg2+/Mg half-cell.
eg2. Cl- won’t be oxidised by HNO3 - it must be reacted with acidified MnO4-.

28
Q

Describe the trend in Group 17 as oxidising agents.

A

As you go down the group, the reduction potentials get less positive. The halogens are less easily reduced and their oxidising power decreases.
The halides’ reducing power will therefore increase as they are more easily oxidised going down the group.

29
Q

What happens to EΘ under non-standard conditions?

A

When we compare the voltage of STD half-cell X with STD hydrogen electrode, we are measuring the EΘ value of X. If T, P or c change, the conditions are non-STD and EΘ changes to E.

30
Q

How does ion concentration affect redox equilibria?

A

eg. Zn2+ + 2e- -> Zn
- if [Zn2+] increases > 1M, E becomes less negative.
- if [Zn2+] decreases < 1M, E becomes more negative.

eg. Fe3+ + e- -> Fe2+
- if [Fe3+] > 1M, E increases
- if [Fe3+] < 1M, E decreases
- if [Fe2+] > 1M, E decreases.
- if [Fe2+] < 1M, E increases.

Increasing [Fe3+] and [Fe2+] will cancel each other out.

Under non-STD conditions, if the EΘ values of the half-reactions involved have a difference >0.30V, the predicted reaction is highly feasible.

31
Q

Do feasibility predictions always work?

A

They only tell us if a reaction is possible and that it won’t occur in the reverse direction, but they don’t tell us anything about the rate.
Sometimes the rate of reaction can be so slow that it seems as if nothing is happening.

eg. Zn in cold water:
- H+ + e- -> 1/2 H2 (EΘ 0.00V)
- Zn2+ + 2e- -> Zn (EΘ -0.76V)

Even considering low [H+], the EΘ values show that the reaction should occur, but it is very slow. Therefore, the rate of reaction determines LACK OF REACTIVITY rather than the EΘ.

32
Q

What does the Nernst equation tell us?

A

The effect of concentration and temperature on Ecell.

33
Q

State the Nernst equation.

A

E = EΘ + RT/zF x ln [oxidised form]/[reduced form]

  • R = gas constant (8.314 J/K/mol)
  • z = number of e- transferred in reaction
34
Q

State the simplified Nernst equation.

A

E = EΘ + 0.059/z x log10 [oxidised form]

  • at STD temperature, R, T and F are constant.
  • [reduced] is the metal concentration which doesn’t change.
35
Q

What does log10[O] tell us?

A

It changes the sign of the second term.

  • if conc. = 1M, log10[O] = 0 and E = EΘ
  • if conc. <1M, log10[O] is - and E < EΘ
  • if conc. >1M, log10[O] is + and E > EΘ
36
Q

Describe ordinary dry cells.

A

They have voltages of 1.5-2.0V. Several are needed to produce enough power, and their voltage drops gradually over time. Cells joined together produce enough voltage but take up space.

37
Q

What must one consider when selecting a cell for a specific function?

A
  • Rechargeability
  • Size + mass
  • Maximum voltage
  • Nature of electrolyte
  • Cost
  • How long the cell can deliver its maximum voltage.
38
Q

Describe rechargeable / secondary / storage cells.

A

Not primary (redox reactions continue until [R] is llow and voltage declines). These are recharged by passing an electric current through, eg. car battery:

  • uses Pb and PbO2 plates immersed in H2SO4 electrolyte.
  • 6 x 2V cells = 12V
  • recharged by the car’s alternator when the engine is running.
39
Q

Describe some improved batteries for electric vehicles.

A

Nickel-cadmium cells - smaller, lower mass, last longer, lower voltage.
Aluminium-air cells - lower mass, higher voltage, more expensive, Al anodes occasionally need replacing.

40
Q

Describe solid state cells.

A

Primary cells developed to produce improved voltage and smaller size and mass.
Used in pacemakers, hearing aids, watches, calculators.
No liquid/paste = no risk of leakage.
Often uses Li or Zn as - pole and iodine / MnO2 / Ag2O as + pole.

41
Q

What are the advantages of other types of cell?

A

Ni-MH - cheap, high energy density, no toxic materials (Ni-Cd).
Li-ion - high energy density, low mass.

Both decrease waste, have low long-term cost and save finite/expensive resources.

42
Q

What is a fuel cell?

A

An electrochemical cell where fuel releases electrons at one electrode and oxygen accepts electrons at the other.

43
Q

How do hydrogen-oxygen fuel cells work?

A
  • H2 and O2 bubbled through two porous platinum-coated electrodes.
  • H2 is oxidised at the negative electrode, H+ ions diffuse through the proton exchange membrane and electrons flow through the external circuit, releasing energy.
  • O2 is reduced to water at the positive electrode.
  • POLE: H2 -> 2H+ + 2e-
    + POLE: 4H+ + O2 + 4e- -> 2H2O
44
Q

What are the advantages and disadvantages of fuel cells?

A

ADVANTAGES:

  • water only product
  • higher energy/g of fuel compared to petrol engines
  • efficient - no moving parts where energy could be lost as thermal; direct transfer from cell to motor.

DISADVANTAGES:

  • expensive (electrodes, membrane)
  • toxic manufacturing byproducts
  • high pressure tanks to store enough fuel, refuelling needed
  • H2 is only produced cheaply by using fossil fuels
  • doesn’t work at cold temperatures - cell ‘freezes’ below zero.
45
Q

Which additional ions are present in an aqueous solution?

A

H+ and OH- from the ionisation of water.

46
Q

Which ions are discharged during the electrolysis of aqueous solutions?

A

It depends on the relative E and concentration of ions.

  • The most easily reduced cation is discharged (eg. NaCl would -> H2 gas).

At the graphite ANODE:

  • the discharge series is followed (increasing ease of oxidation/discharge):
  • sulfate, nitrate, chloride, hydroxide, bromide, iodide.

If conditions are standard, the concentration of the aqueous solution is 1M with respect to the aq. ionic compound.
The concentration of hydrogen and hydroxide ions
in solution is very low, but as long as the difference in electrode potentials is greater than 0.30V, we can be fairly sure that the predictions will be correct.

47
Q

What happens in the electrolysis of conc. Na2SO4 (aq)?

A

ANODE: 4OH- -> O2 + 2H2O + 4e-

- OH- discharged, oxidised -> O2.

48
Q

How are electrolysis products affected by solution concentration?

A

An ion higher in the discharge series may be discharged with preference over one lower if its concentration is much higher than usual.
For this to happen, the E value difference would need to be <0.30V.
For example, dil. NaCl produces both Cl2 and O2 at the anode; proportion of O2 will increase as dilution increases.

49
Q

How do you draw a cell diagram in IUPAC notation?

A

Anode | Reductant, Product || Oxidant, Product | Cathode

  • if either reductant/product is the anode, it is omitted
  • if either oxidant/product is the cathode, it is omitted.
50
Q

Why are salt bridges used in electrolysis cells?

A

To complete the circuit and maintain ionic balance between the half-cells. Ions can use this to travel between solutions so there isn’t a charge buildup in either half-cell.

Chemistry (12 decks)

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