5.2.10: Standard cell potentials

Question 1

How can a simple electrochemical cell be made?

Answer 1

By connecting together two half cells with different electrode potentials.

• One half cell releases electrons.
• The other half cell gains electrons.

Question 2

How can the difference in electrode potential be measured?

Answer 2

With a voltmeter.

Question 3

How are two half cells joined to allow the charge to be carried between each cell via the electrons and the ions?

Answer 3

-The two half cells are joined using a wire and a salt bridge.

Question 4

What can we infer from:
Zn2+(aq) + 2e- Zn(s) E(theta) = -0.76 V
Cu2+ (aq) + 2e- Cu(s) E(theta) = +0.34 V

Answer 4

• Zn half cell has a more -ve value, so has a greater tendency towards the equilibrium shifting left.
• The Zn2+/Zn equilibrium releases electrons into the wire, making zinc the -ve electrode. (oxidation)
• Electrons flow along the wire to the Cu electrode of the Cu2+/Cu half cell. (reduction)

Question 5

What does the reading on the voltmeter measure?

Answer 5

• The potential difference of the cell
• The difference between the electrode potentials of the half cells.
• The bigger the value, the further away from the equilibrium position the reaction moves.
• The reading on the voltmeter can be taken as the cell potential as long as any ions of the same element have a concentration of 1 mol dm-3 or are equimolar.

Question 6

How can the standard cell potential also be calculated?

Answer 6

Ecell = E(+ve terminal) - (E-ve terminal)

Reduction Oxidation

Question 7

How can you determine the feasibility of a reaction?

Answer 7

-By calculating the cell potential for a reaction, using the standard electrode potentials for each half cell determining whether electrons are likely to flow.

Question 8

The half equation for a copper half cell:
Cu2+(aq) + 2e- Cu(s)
From Le Chatelier’s principle, what happens if you increase the concentration of Cu2+ (aq) and how can this affect predictions made?

Answer 8

• The equilibrium opposes the change by moving to the right.
• Electrons are removed from the equilibrium.
• The electrode potential becomes less negative or more positive.
• A change in electrode potential resulting from concentration changes means that predictions made on the basis of standard value may not be valid.

Question 9

Give Some reasons as to why a reaction might not actually take place even if it’s predicted to take place?

Answer 9

• Predictions can be made about the equilibrium position but not about the reaction rate, which may be extremely slow because of a high activation energy.
• The actual conditions used for a reaction may be different from the standard conditions used to measure E(theta) values. This will affect the value of the electrode potential.
• Standard electrode potentials apply to aqueous equilibria and many reactions take place under very different conditions.

Question 10

What is the general working rule to predict whether or not a reaction will take place?

Answer 10

• The larger the difference between E(theta) values, the more likely it is that a reaction will take place.
• If the difference between E(theta) values is less than 0.4 V, then a reaction is unlikely to take place.

Question 11

Electrochemical cells are used as our modern-day cells and batteries. What are the three main types of cells?

Answer 11

• Non-rechargeable cells
• Rechargeable cells
• Fuel cells

Question 12

Describe non-rechargeable cells.

Answer 12

• Provide electrical energy until the chemicals have reacted to such an extent that the voltage falls.
• The cell then falls ‘flat’ and is discarded.

Question 13

Describe rechargeable cells.

Answer 13

• The chemicals in the cell react, providing electrical energy.
• The cell reaction can be reversed during recharging.
• The chemicals in the cell are regenerated and the cell can be used again.

Question 14

Give two common examples of rechargeable cells.

Answer 14

• Nickel and cadmium (Ni-Cad) batteries, used in rechargeable batteries,
• Lithium-iron and lithium-polymer batteries, used in laptops.

Question 15

Describe fuel cells.

Answer 15

The cell reaction uses external supplies of a fuel and an oxidant, which are consumed and need to be continuously supplied.
-The cell will continue to provide electrical energy so long as there is a supply of fuel and oxidant.

Question 16

What are the benefits and drawbacks of lithium batteries which are often found in the home?

Answer 16

Benefits: 
-Provide high levels of battery life. 
Drawbacks:
-Toxicity on being ingested. 
-Rapid discharge of current, which could cause fires and explosions.

Question 17

What are some regulations regarding electrochemical cells?

Answer 17

• Restrictions on the transport of lithium-based batteries.

- limited sales of these batteries to individual consumers in some countries.

Question 18

What are modern fuel cells based on?

Answer 18

-Hydrogen, or hydrogen-rich fuels such as methanol, CH3OH.

Question 19

How does a fuel cell generate energy?

Answer 19

• A fuel cell uses energy from the reaction of a fuel with oxygen to create a voltage.
• The reactants flow in and products flow out while the electrolyte remains in the cell.

Question 20

What must happen for a fuel cell to operate virtually continuously?

Answer 20

• The fuel and oxygen must continuously flow into the cell.

- Fuel cells do not have to be recharged.

Question 21

What are the redox equilibria involved in a fuel cell?

Answer 21

2H2O(l) + 2e- H2 + 2OH- (aq)
E(theta)= -0.83 V, -ve terminal

1/2O2(g) +H2O(l) +2e- 2OH-(aq)
E(theta)= +0.40 V, +ve terminal

The more -ve hydrogen system provides the electrons.

Question 22

Give the overall reaction for a fuel cell and calculate E(cell) given these values:
2H2O(l) + 2e- H2 + 2OH- (aq)
E(theta)= -0.83 V, -ve terminal

1/2O2(g) +H2O(l) +2e- 2OH-(aq)
E(theta)= +0.40 V, +ve terminal

Answer 22

H2(g) + 1/2O2(g) –> H2O(l)
E(cell) =E(+ve terminal) -E(-ve terminal)
=0.4-(-0.83)
=1.23 V