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Chemistry > Electrode Potentials > Flashcards

Electrode Potentials Flashcards

1
Q

Flow of electrons

A

electrons will flow from the more reactive metal (left electrode ) to the less reactive metal (right electrode)

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2
Q

Half Cells

A

When a strip of metal is dipped into a solution of its own ions, an equilibrium is set up.
Cu ⇌ Cu2+ + 2e-

This equilibrium:
- The Cu2+ ions dissolve into the solution. This gives the solution a positive charge.
- The electrons collect into the Copper strip, giving it a negative charge
- This means a potential difference (i.e. a voltage) is established between the two.

  • If there is a LARGE voltage the equilibrium is to the RIGHT
  • If there is a SMALL voltage the equilibrium is to the LEFT
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3
Q

Connecting two half cells

A
  • The two metal rods are connected with wires and a high resistance voltmeter.
  • The two beakers of electrolyte are connected with a salt bridge to complete the circuit. This is a glass tube or porous material soaked in Potassium Nitrate solution.
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4
Q

Why is KNO3 a suitable solution for a salt bridge?

A

KNO3 is unreactive with the electrodes AND the ions are free to move

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5
Q

Voltmeter

A

The voltmeter prevents electrons flowing – this enables the voltage to be measured.

If the volt meter was removed electrons can flow from the left electrode to the right.

If the voltmeter is replaced with an ammeter or a bulb etc., the electrons can flow and a current is produced.

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6
Q

Why might the current produced by a cell fall to zero after some time?

A

All the reactants are used up

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7
Q

What will happen to a cell once the reactants are used up?

A

stops working OR starts to leak

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8
Q

Example: - Copper and Zinc Electrodes

A

Zn(s) → Zn2+(aq) + 2e-
The left electrode is always the negative electrode as electrons are produced there. Oxidation always occurs at the left (negative) electrode.

Cu2+(aq) + 2e- → Cu(aq)
The right electrode is always the positive electrode as electrons are used up there. Reduction always occurs at the right (positive) electrode.

Electrons flow from the negative (left) electrode to the positive (right) electrode.

overall:
Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)

This reaction would continue generating an electrical current until either:
- The solid Zinc rod completely reacted.
- All the Cu2+ ions in solution were used up

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9
Q

Platinum electrodes

A

when there is no solid metal in the reaction, such as when there are metal ions of two different charges in the same solution.

Why is platinum a suitable electrode?
Pt is unreactive AND conducts electricity

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10
Q

The Standard Hydrogen Electrode

A

Key points:
1. H2 gas is pumped in at a pressure of 100kPa
2. The electrolyte contains H+ ions of concentration 1moldm-3
3. There must be a Platinum electrode
4. The whole system must be at a temperature of 298K

The voltage of the Standard Hydrogen Half Cell is defined as ZERO.

This means if another electrode is connected up to the standard hydrogen electrode and voltage between them is measured, the voltage on the meter must be that of the ‘other’ electrode.

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11
Q

standard conditions

A

If the cell connected to the standard hydrogen electrode is also in standard conditions
- Temp = 298K
- All concentrations = 1moldm-3
- All pressured = 100kPa

then the voltage measured can be called:
THE STANDARD ELECTRODE POTENTIAL and be given the symbol Eᶿ

Whenever the phrase “standard electrode potential” is used, one of the electrodes will be the standard hydrogen electrode.

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12
Q

electrochemical series

A
  • All the electrode equations are shown as reductions.
  • The half equation for the standard hydrogen electrode is shown with an electrode potential of zero.
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13
Q

The standard electrode potential of Cu2+/Cu is 0.37 V. Why might the electrode potential of the following cell not be 0.37 V?

A

The concentration of the CuSO4 solution is not 1 mol dm-3.

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14
Q

Conventional Cell Representation

A
  • The right electrode is the positive electrode
  • The right electrode has a more positive standard electrode potential than the left electrode
  • The left electrode is the negative electrode
  • The left electrode has a less positive standard electrode potential than the right electrode
  • Vertical solid lines indicate a phase boundary (i.e. between the solid and aqueous phases).
  • A double vertical line in the middle represents the salt bridge.
  • The species with the highest oxidation state should be written closest to the salt bridge.
  • The order of the electrodes is important too. In the example above the Zinc electrode is on the left, and so it is also on the left of the cell representation
  • add comma between those in same phase
  • add Pt if no metal
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15
Q

cell representation of the standard hydrogen half cell

A

Pt(s)│H2(g) │H+(aq) ‖

The standard hydrogen electrode is always written on the left.

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16
Q

The conventional cell representation of a solution containing Fe2+ and Fe3+ ions

A

‖ Fe3+(aq), Fe2+(aq) │Pt(s)

17
Q

Calculations using Standard Electrode Potentials

A

All species on the left of the arrow are oxidising agents (as they can only gain electrons)

All species on the right of the arrow are reducing agents (as they can only donate electrons)

“SOWR”

18
Q

Why will Ag+(aq) ions react with Li(s)?

A

Eᶿ Ag+(aq) > Eᶿ Li(s) AND Ag+(aq) is a stronger oxidising agent than Li(s)

19
Q

What will be observed when Ag+(aq) ions react with Li(s)?

A

Solid Ag forms

20
Q

What happens when Li+ is added to Ag?

A

Eᶿ Ag(s) > Eᶿ Li+(aq) AND Li+ cannot oxidise Ag

21
Q

Identify a chemical that reacts with Cu(s).

A

Ag+(aq) AND Eᶿ Ag+(aq) > Eᶿ Cu(s)

22
Q

Working out the voltage of a cell

A

Ecell = more positive – least positive

23
Q

construct the equation for the feasible reaction

A
24
Q

Cell discharge and recharge

A

When the Ecell value is positive, this means the reaction is feasible and the cell discharges – this means it produces a current.
e.g.
Ag+(aq) + Li(s) → Ag(s) + Li+(s)

Cell recharge
If a reaction is reversible, the cell can be recharged by plugging it into the mains. The reverse reaction will occur when the cell is recharged.
e.g.
Ag(s) + Li+(s) → Ag+(aq) + Li(s)

25
Q

Give an environmental advantage of using rechargeable cells.

A

Metals are reused

26
Q

Give an environmental disadvantage of using rechargeable cells.

A

Mains electricity is used to recharge, which may come from combusting fossil fuels, which releases CO2(g)

27
Q

Why are conditions so important?

A

Breaking endo = ↑ temp
making exo = ↓ temp

28
Q

The cell made of the systems Mg/Mg2+ and Fe2+/Fe under standard conditions has an e.m.f. of 1.93 V.

Deduce how the e.m.f. of the cell Mg/Mg2+ and Fe2+/Fe changes when the concentration of Mg2+ is decreased. Explain your answer. (3 marks)

A
29
Q

In which direction would the electrons flow in the above cell?

A

From right to left
The Cu2+ is more concentrated on the left, so reduction on Cu2+ is more likely to happen on the left (Cu2+ + 2e- à Cu)
The left electrode is the positive electrode, so the right electrode is the negative electrode

30
Q

Single Use Batteries

A

electrochemical reaction taking place inside the cell is irreversible. Once the reactants have been used up then it cannot be used again.

31
Q

Rechargeable Batteries

A
32
Q

Advantages of using fuel cells for energy instead of fossil fuels

A
  • Major advantage; Greater efficiency than burning hydrogen in a combustion engine
  • Less-polluting as water is the only product
33
Q

Disadvantages of using fuel cells for energy instead of fossil fuels.

A
  • H2 is difficult to store
  • Fossil fuels are combusted to produce the hydrogen, which releases carbon dioxide
34
Q

Advantages of fuel cells compared to other types of cell.

A
  • Voltage is constant, as fuel and oxygen is supplied constantly so concentrations of reactants remain constant.
35
Q

The Hydrogen-oxygen fuel cell: ACIDIC

A

At the anode (negative electrode)
H2(g) ⇌ 2H+(aq) + 2e-
At the cathode (positive electrode)
½ O2(g) + 2H+(aq) + 2e- ⇌ H2O(l)

Overall: H2(g) + ½ O2(g) → H2O(l) E = 1.23V

36
Q

The Hydrogen-oxygen fuel cell: ALKALI

A

At the anode (negative electrode)
H2(g) + 2OH-(aq) ⇌ 2H2O(l) + 2e- = oxidation
At the cathode (positive electrode)
½ O2(g) + H2O(l) + 2e- ⇌ 2OH-(aq) =reduction

Overall: H2(g) + ½ O2(g) → H2O(l) E = 1.23V

Chemistry (30 decks)

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