Chapter 23 - Redox and electrode potentials

Question 1

What is an oxidising agent?

Answer 1

An oxidising agent takes electrons from the species being oxidised, getting reduced itself

Question 2

What is a reducing agent?

Answer 2

A reducing agent adds electrons to the species being reduced, getting oxidised =

Question 3

What is the oxidising agent and reducing agent in this equation?

2Ag⁺(aq) + Cu(s) → Cu²⁺(aq) + 2Ag(s)

Answer 3

oxidising agent - Ag⁺ (Ag⁺ oxidises Cu)

reducing agent - Cu

Question 4

What are the steps in writing a redox equation from half equations?

Answer 4

step 1: Balance electrons

step 2: Add + cancel electrons

step 3: finally cancel any species that are on both sides of the equation

Question 5

e. g. Redox reaction between H₂O₂ and Cr³⁺ ions have the following half equations, write the overall redox eqution
reduction: H₂O₂ + 2e^- → 2OH^-
oxidation: Cr³⁺ + 8OH^- → CrO₄^8- + 4H₂O + 3e^-

Answer 5

reduction x 3 to get 6e^-

oxidation x 2 to get 6e^-

3H₂O₂ + 6e^- → 6OH^-

2Cr³⁺+16OH^- → 2CrO₄^2- + 8H₂O + 6e^-

cancel electrons

2Cr³⁺ + H₂O₂ + 16OH^- → 6OH^- +2CrO₄^2- + 8H₂O

cancel any species on both sides

2Cr³⁺ + 3H₂O₂ + 10OH^- → 2CrO₄^2- + 8H₂O

Question 6

What are the steps in writing a redox reaction from oxidation numbers?

Answer 6

step 1: Summarise the information in question

step 2: Assign oxidation numbers to identify the atoms that change oxidation number

step 3: Balance only the species that contain the elements that have changed oxidation number

step 4: Balance only remaining atoms

Question 7

e.g. construct overall redox equation for:

S + HNO₃ → H₂SO₄ + NO₂ + H₂O

Answer 7

Assign numbers:

S + HNO₃ → H₂SO₄ + NO₂ + H₂O

0 +5. +6. +4

S oxidation change = +6

N oxidation change = -1

Balance only the species that contain the elements that have changed oxidation number:

To match increase of +6 for sulphur, you need a total decrease of -6 from nitrogen

So HNO₃ and NO₂ also need to be multiplied by 6 to give 6x-1 = -6

S + 6HNO₃ → H₂SO₄ + 6NO₂ + H₂O

Balance any remaining atoms:

S + 6HNO₃ → H₂SO₄ + 6NO₂ + 2H₂O

Question 8

How is an imbalance of oxygen atoms on one side of an equation balanced?

Answer 8

Put H₂O’s on the other side of reaction

Question 9

How is an imbalance of hydrogen atoms balanced in a redox reaction?

Answer 9

Put H⁺ ions on the opposite side of reaction

Question 10

How do you predict products of a redox reaction?

Answer 10

• Write half equation
• assign oxidation numbers (and the change in oxidation numbers)
• Balance the electrons
• balance any remaining atoms and predict any firther species

(unlikely to add species other than H₂O, H⁺ and OH^-) (H₂O often formed in aqueous redox reactions)

• ensure that both sides of the equations are balanced by charge

Question 11

Describe the procedure of a manganate (VII) titration. (redox titration)

Answer 11

• MnO₄^- ions are reduced and so the other chemical must be a reducing agent
  1. ) Stndard solution of potassium manganate (VII), KMnO₄ is added to the burette.
  2. ) Using a pippette add measured volume of solution being analysed to conical flask. Excess of dilute sulphuric acid is also added to provide the H⁺(aq) ions required for the reaction of MnO₄^-(aq) ions. The reaction is self-indicating, so no indicator is needed.
  3. ) Manganate (VII) solution is decolourised as it’s added - the end point of the titration is marked by permanent pink colour (indicating excess of MnO₄^-)
  4. ) Repeat until concordant results are formed (+- 0.1cm³)

Question 12

What are 2 examples of reducing agents to analyse in a manganate (VII) titration?

Answer 12

• Iron (II) ions - Fe²⁺(aq)
• ethandioic acid - (COOH)₂(aq)

reduce MnO₄^- to Mn²⁺

Question 13

What is the manganate half equation?

Answer 13

MnO₄^-(aq) + 8H⁺(aq) + 5e^- → Mn²⁺(aq) + 4H₂O(l)

Question 14

What are the steps in solving:

21.60 cm³ of 0.015 moldm^-3 KMnO₄ is added to 30 cm³ of a solution containing an unknown amount of Iron(II) ions and at this point the solution changes colour from colourless to purple. What was the concentration of Iron(II) in the solution before the titration

Answer 14

1.) Find the number of moles of MnO₄^- that reaated:

n = 21.60/1000 x 0.015 = 3.24 x 10^-4 moles

2.) Use Redox equation to find number of Fe²⁺ present

MnO₄ + 8H⁺ + 5Fe → Mn²⁺ + 4H₂O + 5Fe³⁺

For every 1 mol of MnO₄^- reacted, 5 moles of Fe react,

so 3.24 x 10^-4 moles x 5 = 1.62 x 10^-3 mol

Find concentration of solution:

c = 1.62 x 10^-3/0.03

0.054 moldm^-3

Question 15

What other oxidising agent could be used in a redox titration other than KMnO₄?

Answer 15

H+/Cr₂O₇^2-

Question 16

What’s reduced and what’s oxidised on iodine/thiosulphate redox reaction?

Answer 16

• Thiosulphate ionns, S₂O₃^2-(aq) are oxidised
• Iodine, I₂, are reduced

Question 17

What is the iodine/sodium thiosulfate reaction equation?

Answer 17

S₂O₃^2-(aq) + I₂(aq) → 2I^-(aq) + S₄O₆^2-(aq)

Question 18

Describe the process of an iodine/thiosulphate redox titration (to analyse oxidising agents)

Answer 18

1. ) Add standard solution of Na₂S₂O₃ to the burette
2. ) Prepare a solution of oxidising agent to be analysed. Add solution to a conical flask using a pipette. Add an excess of potassium iodide - reacts with oxidising agent to form iodine, turning the solution a yellow-brown.
3. ) Titrate solution with Na₂S₂O₃(aq). Iodine reduced back to I^- ions and brown colour fades. Starch is added and when the end point is approached and the iodine colour fades to become a pale straw. Starch added forms blue-black solution which turns colourless at the end point.

Question 19

What are 2 examples for the analysis of oxidising agents using Iodine/thiosulphate titrations?

Answer 19

• Chlorate (I) ions, ClO^-(aq)
• Copper (II) ions, Cu²⁺(aq)
• Any oxidising agent that can oxidise I^- ions to I₂.

Question 20

Thiosulphate example:

A sample of aqueous bromine solution is mixed with an equal volume of excess potassium iodide and starch, the bromine acts as an oxidising agent

14.5 cm³ of 0.020 moldm^-3 Na₂S₂O₃ is then added to 15cm³ of the new solution and at this point the solution changes colour from inky blue to straw yellow

What was the concentration of the bromide solution before the potassium iodide was added?

Answer 20

Write the half equations for the reduction of the bromine and the oxidation of the iodide ions:

Br₂(aq) +2e^- → 2Br-

2I^- → 2e^- + I₂

Construct the full redox equation:

2I^- + Br₂ → I₂ + 2Br^-

Find the number of moles of S₂O₄^2- that reacted in the titration:

0.02 moldm^-3 x 14.5/1000 = 2.9 x 10^-4 mol

Use the titration equation to find the number of I₂ present:

2S₂O₃^-(aq) + I₂(aq) → S₄O₆^2-(aq) + 2I^-(aq)

2. : 1
3. 45 x 10^-4 mol of iodine

Use the recation equation for bromine and iodide ions to find the moles of bromine molecules which made the iodine:

2I^- + Br₂ → I₂ + 2Br^-

^{1. : 1}

Mol of bromine = 1.45 x 10^-4 mol

Work out the concentration:

c = 1.45 x 10^-4 mol/ 0.015 dm³

= 9.7 x 10^-3 moldm^-3

Find concentration of bromine ions in the original soluion:

0.0194 moldm^-3

Question 21

What’s a voltaic cell and what do you need to make one?

Answer 21

• Type of electrochemical cell that converts chemical energy into electrical energy
• You need chemical reactions that transfer electrons from one species to another (redox reactions) as electrical energy results from movement of electrons

Question 22

How can a voltaic cell be made by connecting 2 half-cells?

half cells - contains chemical species in a redox reaction

Answer 22

Allows electrons to flow

Question 23

Draw a metal/metal ion half cell for Zn²⁺(aq)/Zn(s)

Answer 23

Question 24

Draw an ion/ion half cell for Fe (3+ and 2+)

Answer 24

Question 25

Why is an inert platinum electrode put in the ion half cell?

Answer 25

No metal to transport electrons either into or out of the ion/ion half cell

Question 26

When two metal/metal ion half cells are connected, which is oxidised and which is reduced?

Answer 26

More reactive metal is oxidised

Less reactive metal is reduced

Question 27

Define standard electrode potential (E - standard conditions)

Answer 27

emf of a half cell connected to a standard hydrogen half-cell under standard conditions of 298K, with solution concentrations of 1moldm^-3 and 100KPa pressure

Question 28

Draw the standard hydrogen electrode

Answer 28

Question 29

Whats the standard electrode potential of a standard hydrogen electrode?

Answer 29

0v

Question 30

What does the standard electrode potential tell you?

Answer 30

relative tendency to be reduced and gain electrons compared with the hydrogen half-cell

Question 31

How do you measure a standard electrode potential?

Answer 31

connect half cells to a standard hydrogen electrode

• 2 electrodes connected by a wire to control flow of electrons
• 2 solutions connected with a salt bridge which allows ions to flow (solution of electrode that doesn’t react with either solution e.g. filter paper soaked in aqueous potassium nitrate KNO₃(aq)

Question 32

Draw the experimental set up for measuring the standard electrode potential for a copper half-cell

Answer 32

Question 33

What does a more negative E(standard conditions) value signify?

Answer 33

• greater the tendency to lose electrons and undergo oxidation
• less tendency to gain electrons and under go reduction
• greater the reactivity in a metal (in losing electrons)

Question 34

What does a more positive E(standard potential) value signify?

Answer 34

• greater the tendency to gain electrons undergo reduction
• less tendency to lose electrons and undergo oxidation
• greater the reactivity of a non-metal (in gaining electrons)

Question 35

What E(standard conditions) values do metals and non-metals tend to have?

Answer 35

• metals tend to have negative E(standard conditions) values
• Non-metals tend to have positive E(standard conditions) values

Question 36

How do you measure standard cell potentials?

Answer 36

1.) Prepare 2 standard half-cells. For metal/metal ion half cell, the metal ion must have a concentration of 1moldm^-3. Temp must = 298K

ion/ion half cell must have some concentration metal ions in solution

half cells containing gas must be at 100 KPa in contact with ionic concentration of 1moldm^-3. Must be an inert electrode - usually platinum.

2. ) Connect the metal electrodes of the half cells to a voltmeter using wires
3. ) Prepare a salt bridge by soaking strip of filter paper in a saturated aqueous solution of KNO₃.
4. ) Connect 2 solutions of the half cells with a salt bridge

Question 37

Draw a standard cell made from Zn²⁺(aq)/Zn(s) and Cu²⁺(aq)/Cu(s) half cells

Answer 37

Question 38

If standard cell potential is 1.1.V and the copper half cell has the more psoitive E(standard consitions) and zinc half cell is more negative. What will happen?

Answer 38

Copper half cell has greater tendency to undergo reduction and the zinc half cell has greater tendency to undergo oxidation. Electrons flow along the wire from the more negative zinc half-cell to the less negative copper. zinc electrode is negative and copper electrode is positive

Question 39

Write the overall cell equation for the standard cell of Zn²⁺(aq)/Zn(s) and Cu²⁺(aq)/Cu(s) half cells

Answer 39

reduction: Cu²⁺(aq) +2e^- → Cu(s) (positive electrode)
oxidation: Zn(s) → Zn²⁺(aq) + 2e^- (negative electrode)

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

Question 40

How do you calculate the standard cell potential from standard electrode potentials?

Answer 40

E(standard conditions) = E(standard conditions)positive electrode - E(standard conditions)negative electrode

e.g. for zinc-copper cell E(standard conditions)cell = 0.34 -(-0.76) = 1.10V

Question 41

How do you know if a cell is feasible?

Answer 41

provided that the redox system of oxidising agent has a more positive standard electrode potential than the redox system of the reducig agent, a reaction should take place between them.

Question 42

What do you need to do to predict if a reaction will happen and which direction it will go in?

Answer 42

1. ) Find the 2 half equations for the redox reaction, and write them both out as reduction reactions.
2. ) Observe which half equations has the more negative electrode potential going in the backwards direction (oxidation) and the half-equation with the more positive electrode potential going in the forwards direction (reduction)
3. ) Write out the half-equation with the more negative electrode potential going in the backwards direction direction (oxidation) and the half equation with the more positive electrode potential going in the forwards direction (reduction).
4. ) Combine the two half equations and write out a full redox equation. If this id the feasible direction it will give a positive E(standard conditions) value.

Question 43

*Predict the direction of the reaction when Zn/Zn²⁺ half cell is connected to a Cu²⁺/Cu half cell

Answer 43

Question 44

What are the limitations of predicting feasibility and direction of reaction using E(standard conditions)?

Answer 44

• Conditions are not standard (change of concentration)
• Reaction rate (kinetics). May be so slow that the reaction might not appear to happen. If a reaction has a high activation energy - may stop it happening

Question 45

What happens if the concentrations of Zn²⁺ is >1moldm^-3 and <1moldm^-3 in the Zn²⁺/Zn redox system:

Zn²⁺(aq) +2e^- ⇔ Zn(s) E(standard conditions)=-0.76V

Answer 45

>1moldm^-3: equilibrium shifts to the right, removing electrons from the system, making the electrode potential less negative

<1moldm^-3: equilibrium shifts to left, increasing electrons in the system and making electrode potential more negative

Question 46

What 3 main types can baterries be divided into?

Answer 46

primary, secondary and fuel cells

Question 47

What are the main features of primary cells?

Answer 47

• Non-rechargable (for one use only)
• electrical energy produced by oxidation and reduction at electrodes.
• Reactions can’t be reversed (chemicals will be used up)
• cell will be discarded or recycled when chemicals are used up, causing voltage to fall and battery goes flat
• Most are alkaline based on zinc and managanese dioxide

Question 48

What uses do primary cells have?

Answer 48

In low-current, long storage devices such as wall clocks and smoke detectors

Question 49

What are differences between secondary cells and primary cells?

Answer 49

• rechargable

cell reaction producing energy can be reversed during recharging. Chemicals in cell are regenerated

Question 50

What are common examples of secondary cells?

Answer 50

• Lead-acid batteries used in car batteries
• Nickel-cadmium, NiCd, cells and Nickel-metal hydride, NiMH - the cylindrical batteries used in radios, torches and so on.
• Lithium-ion and lithium-ion polymer cells used in our modern appliances - laptops, tablets, cameras and also being developed for cars

Question 51

State the key characteristics of fuel cells.

Answer 51

• Uses energy from the reaction of a fuel with oxygen to create a voltage
• fuel + oxygen flow into the fuel cell and the products flow out. The electrolyte remains in the cell.
• fuel cells operate (provide electrical energy) continuously provided that the fuel and oxygen are supplied into the cell.
• don’t have to be recharged

Question 52

What is the most common fuel for a fuel cell and why?

Answer 52

Hydrogen as it produces no CO₂ during combustion (H₂O) is the only product

Question 53

Give 2 advantages and 2 disadvantages of using hydrogen fuel cells (over fossil fuels)?

Answer 53

ADV: - Only H₂O formed (no pollutants produced e.g. CO₂)

• greater efficiency

DSV: - H₂ difficult to store

• H₂ difficult to manufactire
• Limited life cycle of H₂ absorber

Question 54

If the standard electrode postential is higher than another is it a better or worse oxidising agent?

Answer 54

Better

Question 55

Which half-cell is the anode and which is the cathode?

Answer 55

Most positive electrode potential half-cell is the positive electrode - cathode

Most negative electrode potential half-cell is the negative electrode - anode

Question 56

Which electrode does oxidation and reduction occur?

Answer 56

oxidation - anode

Reduction - cathode

Question 57

When water is added to Cu²⁺(aq)/Cu(s) half cell, connceted to a Ag⁺(aq)/Ag(s) half cell, why does the cell potential change?

Answer 57

[Cu²⁺] decreases shifting equilibrium to the left, so more electrons released by Cu - making electrode potential more negative - so cell potential gets larger

Question 58

How can hydrogen be stored (as a fuel for vehicles)?

Answer 58

• As a liquid under pressure
• On the surface of a solid material

Question 59

Whats the difference between an acidic and alkaline hydrogen fuel cell?

Answer 59

In an acidic hydrogen fuel cell, the electrolyte is acidic and facilitates the flow of H⁺ ions - connecting the two terminals

In an alkaline hydrogen fuel cell, the electrolyte is alkaline and facilitates flow of OH^- ions - connecting the two terminals

Question 60

What’s the oxidation and reduction half cell equations for alakline hydrogen fuel cell?

Also give the overall equation

Answer 60

oxidation: H₂(g) + 2OH^-(aq) → 2H₂O(l) + 2e^-
reduction: 1/2O₂(g) + H₂O + 2e^-→ 2OH^-(aq)

Overall: H₂(g) + 1/2O₂(g) → H₂O(l)

Question 61

What’s the oxidation and reduction half cell equations for an acidic hydrogen fuel cell?

Also give the overall equation

Answer 61

Oxidation: H₂(g) → 2H⁺(aq) + 2e^-

reduction: 1/2O₂(g) + 2H⁺(aq) + 2e^- → H₂O(l)

Overall: H2(g) + 1/2O2(g) → H2O(l)