Electrochemistry

Question 1

Purpose of Electrolysis

Answer 1

It is often used to extract metals that are high in the reactivity series. These metals cannot be extracted by heating their ores with carbon. Electrolysis is also used to produce non-metals such as chlorine and purify some metals.

Question 2

Electrolysis is what type of reaction?

Answer 2

Redox

Question 3

The mass of the substance produced at the electrode during electrolysis is proportional to

Answer 3

1. the time over which a constant electric current (d.c) passes.
2. the strength of the electric current.

Question 4

Faraday

Answer 4

F = charge/1 mole of e-
One Faraday is the quantity of electric charge carried by 1 mole of electrons or 1 mole of singly charged ions. Its value is 96500 C/mol. (Relationship between faraday and the Avogadro constant is F=Le)
i.e. Ag+ + e- —-> Ag (One Faraday of electricity 96500C is required to deposit 1 mole of Silver)
i.e. 2Cl- —-> Cl2 + 2e- (Two Faraday of electricity 2x96500 C to deposit 1 mole of Copper)

Question 5

Value of F (Faraday) can be used to calculate?

Answer 5

1. the mass of substance deposited at an electrode
2. the volume of gas produced at an electrode
   (Worked example 1 & 2 pg no. 408)

Question 6

STEPS TO FIND MASS OR VOLUME USING FARADAY’S CoNSTANT:

Answer 6

1. Write the half-equation.
2. Find the number of coulombs required to deposit 1 mole of product at the electrode (F x no. of electrons in half equation).
3. Calculate the charge transferred during the electrolysis(Q = I x t).
4. Calculate the Mass (mass = Charge calculated in step 3 divided by amount of Faraday in C mol-1 in step 2 and finally multiply by Mr) or Volume (volume = Charge calculated in step 3 divided by amount of Faraday in C mol-1 in step 2 and finally multiply by 24.0 dm3) by simple proportions.

Question 7

Value of L (Avogadro constant) can be found using

Answer 7

L= (charge on 1 mole of electrons) ÷ (charge on 1 electron)
(Practical Activity 20.1 pg no. 409 + Texbook Qs 8, 9.) —->Difficult so review from time to time.

Question 8

Electrode potential (E) vs Standard cell potential

Answer 8

Electrode potential, E: the voltage measured for a half-cell compared with another half-cell under standard conditions.
Standard cell potential: the difference in standard electrode potential between two specified half cells under standard conditions.

Question 9

When is redox equilibrium established

Answer 9

when the rate of electron gain equals the rate of electron loss

Question 10

Standard conditions for electrolysis

Answer 10

101 KPa/ 1 atm, 1.00 mol/dm3 and 298K

Question 11

Why is platinum black used as an electrode?

Answer 11

The platinum black is finely divided platinum. This allows close contact of hydrogen gas and H+ ions is established quickly. The platinum electrode is inert so it does not take part in the reaction.

Question 12

What do Electrode potential values indicate?

Answer 12

Electrode potential values give us an indication of how easy it is to reduce a substance.
» The more positive (or less negative) the electrode potential, the easier it is to reduce the ions on the left. So the metal on the right is relatively poor reducing agent. e.g. Ag+ + e- <——> Ag
» The more negative (or less positive) the electrode potential, the more difficult it is to reduce the ions on the left. So the metal on the right is relatively reactive and is relatively good reducing agent. e.g. Zn2+ + 2e- <—-> Zn

Question 13

Electron flow in the electrolytic circuit

Answer 13

Electrons flow round the external circuit from the metal with the more negative (or less positive) electrode potential to the metal with less negative (or more positive) electrode potential

Question 14

Half-equations for reactions taking place at the half-cells.
⁍ Br2/2Br-
⁍ O2 + H20/OH-
⁍ VO2+ + H2O/VO2 + —>whole charge of Plus 1

Answer 14

• 2Br–(aq) → Br2(l) + 2e–
• O2(g) + 4H+(aq) + 4e- —> 2H2O(l)
• VO2+(aq) + 2H+(aq) + e- —> VO2+(aq) + H2O(l).

Question 15

Purpose of salt bridge

Answer 15

a salt bridge completes the electrical circuit allowing the movement of ions between the two half-cells so that the ionic balance is maintained. It does not allow the movement of electrons. A salt bridge can be made from a strip of filter paper (or other inert porous material) soaked in a saturated solution of potassium nitrate.

Question 16

Why aqueous silver nitrate is not used in a salt bridge when connecting half-cell containing Zn and 1.00 mol ZnCl2(aq) to another half-cell?

Answer 16

Silver ions (Ag+) will react with the Chloride ions (Cl-) to form a precipitate therefore, consuming the free ions of Silver ions, impairing the flow of current

Question 17

What do the positive and negative terminal indicate?

Answer 17

Reduction takes place at the positive terminal and Oxidation takes place at the negative terminal.

Question 18

Calculating standard cell potential

Answer 18

Sum of both the cell potential (or more positive − less positive)

Question 19

The tendency of the species on left and right side of the equation to reduce and oxidise, determined by their respective electrode potentials in a standard cell

Answer 19

For each half-equation, the more oxidised form is on the left and the more reduced form is on the right.
» The more positive the value of Eo, the greater tendency for the half-equation to proceed in the forward direction.
» The less positive the value of Eo, the greater the tendency for the half-equation to proceed in the reverse direction.

Question 20

Feasible

Answer 20

A reaction is likely to take place if E○cell has a positive value.

Question 21

When is the redox reaction in electrolysis likely to occur?

Answer 21

A redox reaction occurs in the direction where the better oxidising agent reacts with the better reducing agent.

Question 22

Two ways to predict if a reaction is likely to occur

Answer 22

1. Using E○ values (Identify the stronger reducing and oxidising agent and predict if the reaction takes place)
2. Relative voltage of the half-cells (Write the equation for reduction first and add the electrode potential which is more positive. Then, write the oxidation equation by reversing it with e- released on right side and add the electrode potential by switching the sign to negative as it’s oxidation)

1. Worked example 3, 4 & 5 pg. 420,421& 422.
2. Worked example 6 & 7 pg. 422

Question 23

As the values of E○ for reduction reactions gets more negative

Answer 23

◘ The species on the left of the equation becomes weaker oxidising agents. They accept electrons less readily.
◘ The species on the right of the equation become stronger reducing agents. They release electrons more readily.

Question 24

What trend does Group 17 from F2 to I2 follow based on their E○ values?

Answer 24

As we go down the group 17, the oxidising ability of the halogen decreases and the ability of the halide ions to act as reducing agents increases.

Question 25

Effect of changing the concentration or temperature of half-cell X

Answer 25

The electrode potential also changes and we use the symbol E instead of E○ for the electrode potential.
i.e. Zn 2 + ( aq ) + 2 e - → Zn ( s )
⁍ If [Zn 2 +] is greater than 1.00 moldm-3, the value of E becomes less negative / more positive (for example -0.61 V).
⁍ If [Zn 2 +] is less than 1.00 moldm-3, the value of E becomes more negative / less positive (for example, 0.80 V).
(If we increase the concentration of the species on the left of the equation, the position of equilibrium will shift to the right. So the value of E becomes more positive / less negative.)

Question 26

The value above which the reaction is likely to occur under non-standard conditions

Answer 26

0.30 V

Question 27

Nernst equation

Answer 27

➊ E = E○ + 0.059/z * log10 [oxidised species] / [reduced species]
➋ E = E○ – [RT/zF] * log 10 [oxidised species] / [reduced species]—> (when temperature is not 298K)

Note that the concentration of the metal does not change i.e. Ag(s). So, the ratio [oxidised species] / [reduced species] can be written as [oxidised species] i.e. [Cu 2+ (aq) moldm-3]

Question 28

How does changing log 10 [oxidised form] affect the value of E relative to E○?

Answer 28

► If the concentration is 1 moldm−3, log10 [oxidised form] is 0 and E = E○.
► If the concentration is less than 1 moldm−3, log10 [oxidised form] is negative and E is less positive than E○.
► If the concentration is greater than 1 moldm−3, log10 [oxdised form] is positive and E is more positive than E○.

Question 29

What determines the feasibility when E○ value does not work?

Answer 29

The rate of reaction rather than the value of E○ which is determinesthe lack of reactivity.

Question 30

Three things which decide the product at an electrode

Answer 30

1. Is the electrolyte molten or an aqueous solution?
2. Are the discharge potentials of the competing ions, e.g. OH− and Cl−, in the electrochemical series (discharge series) similar or significantly different?
3. Is the concentration of the relevant ions high or low?

Question 31

Electrolysis products and solution concentration

Answer 31

When a concentrated solution of sodium chloride is electrolysed, chloride ions are discharged at the anode in preference to hydroxide ions. This is because chloride ions are present in a much higher concentration than hydroxide ions. The chloride ions fall below the hydroxide ions in the discharge series. But what happens when we electrolyse an extremely dilute solution of sodium chloride? We find that oxygen, rather than chlorine, is formed at the anode. This is because the relatively lower concentration of Cl− ions allows OH− ions to fall below Cl− ions in the discharge series. In reality, the electrolysis of a dilute aqueous solution of sodium chloride gives a mixture of chlorine and oxygen at the anode. The proportion of oxygen increases, the more dilute the solution.