(4) redox reactions

UHS Tutoring

(4) Redox Reactions

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UHS Tutoring Students learn to: Students:

4. Oxidation-reduction reactions are increasingly important as a source of energy

A. Explain the displacement of metals from solution in terms of transfer of electrons

Perform a first-hand investigation to identify the conditions under which a galvanic cell is produced

Perform a first-hand investigation and gather first-hand information to measure the difference in potential of different combinations of metals in an electrolyte solution

Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following:

Button cell

Fuel cell

Vanadium redox cell

Lithium cell

Liquid junction photovoltaic device (eg the Gratzel cell)

In terms of:

Chemistry

Cost and practicality

Impact on society

Environmental impact

Solve problems and analyse information to calculate the potential requirement of named electrochemical processes using tables of standard potentials and half-equations

B. Identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals

C. Account for changes in the oxidation state of species in terms of their loss or gain of electrons

D. Describe and explain galvanic cells in terms of oxidation/reduction reactions

E. Outline the construction of galvanic cells and trace the direction of electron flow

F. Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells

HSC CHEMISTRY Lesson 5: Redox reactions

Copyright © United Hemispheres Services 2014 All rights reserved. No part of this book may be reproduced by any person or entity, in any form or by any means, electronic or mechanical, including photocopying without the prior written permission of United Hemispheres Services.

What are redox reactions?

Redox reactions are a type of displacement reactions (as you will see below). They are an

important class of reactions in chemistry and give some students hard time.

You need to make sure that you have understood a couple of essential concepts before you

get into Redox reactions, some of these concepts include:

a. Chemical reactions and electron transfer

b. Assigning oxidation numbers (Important for knowing which species loose and

which gain electrons when redox reactions happen. You will be asked to

identify that in the HSC)

c. Writing net ionic equations (Because redox reactions are essentially ionic

equations).

First let us ask: What is a net ionic equation?

A net ionic equation is the equation that takes into consideration the ions that participate in

the reaction (and nothing else).

Can you remember what spectator ions are?

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- Steps involved in Redox reactions

Balance the equation Assign oxidation numbers

Write net ionic equation Notice changes in oxidation numbers

Write Redox half reactions Balancing Redox reactions

Calculate cell potential

1 2

4 3

6 5

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Balance the equation

As learnt in year 11.

Assign Oxidation Numbers

There are 6 main rules you use when assigning oxidation numbers. The rules are outlined in the table

below:

Rule

1 A

Neutral element or compound

H2

MgBr2

NaNO3

ZERO

1B

Ions

PO43-

= -3

NH41+

= +1

Sum = charge

on that ion

2A In compounds, Group 1A Elements in Group 1 +1

2B In compounds, Group 2A Elements in Group 2 +2

3 In compounds, Fluorine ALWAYS +1

4 In compounds, Hydrogen (Eg.). HCl +1 When hydrogen is

bound to a metal,

it is a -1

5 In most compounds, Oxygen

-2 Exception in

peroxides like H2O2

it is -1.

6 In 2-element compounds with

metals

Group 7A = -1

Group 6A = -2

Group 5A = -3

When assigning oxidation numbers, we follow these rules in PRIORITY ORDER. Which means that if

the first rule for example is sufficient to assign the oxidation number, we don’t have to look at the rule

after it.

1

2



+

What would be the oxidation number of SO4 in Na2SO4?

Step 1: See if rule 1A applies.

Yes.

Rule number 1 applies. Therefore the total charge of the molecule must ZERO.

Is this enough to work out the oxidation number of SO4?

No. So keep going.

Step 2: See if rule 1B applies.

No.

We do not have any charges displayed on our compound.


No. So keep going.

Step 3: See if rule 2A applies

Yes.

Rule 2A applies to Na because it is a metal from group 2 on the periodic table and it must

have the charge.


Yes. So stop here and do the maths.

We know now that Na has a charge of +1 so Na2 has a charge of +2.

We know that the total charge is ZERO.

(+2) + (SO4) = 0

(SO4) = 0 - (+2)

(SO4) = -2

WORKED EXAMPLE



1) Assign an oxidation number for Chlorine in: KClO4

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2) Assign an oxidation number for Sulphur in: K2SO4

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The oxidation number can be a fraction. It does not have to be an integer

TRY IT YOURSELF



Write net ionic equations

Write the complete ionic equation and the net ionic equation for the following reaction:

Ca(s) + 2HBr(aq) CaBr2(aq) + H2(g)

Solution

Step 1: You break down all the aqueous solution compounds into individual ions (Remember: you do not breakdown solids or gases). Ca(s) + 2H+ + 2Br- Ca2+ + 2Br- + H2(g)………………………………… Complete ionic equation

Step 2: Just get rid of the spectator ions

Ca(s) + 2H+ + 2Br- Ca2+ + 2Br- + H2(g)

Ca(s) + 2H+ Ca2+ + H2(g) ………………………..……………………………… Net ionic equation

WORKED EXAMPLE

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Write the net ionic equation for the following reaction:

1. BaBr2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaBr (aq)

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2. CuSO4(aq) + Zn(s) ZnSO4(aq) + Cu(s)

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3. Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate. ……………………………………………………………………………………………………………………………………

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TRY IT YOURSELF



Let us now look at each dot point in details and make sure we understand the exam requirements.

A. explain the displacement of metals from solution in terms of transfer of electrons

When we say displacement of metals from solution, we mean that one metal is kicking out another

metal from the solution and taking its place.

You notice that there are two metals, each immersed in a liquid. The first metal (to the left)

is losing electrons (e-) (being converted from Zn(s)0 into Zn2+ ions) and the second metal (to

the right) is gaining those electrons (being converted from Cu2+ ions into Cu(s)0).

When the Solid Zinc (to the left) starts to convert into Zn2+ ions (dissolved in water – Aqueous),

the weight of Solid Zinc will decrease. This means that it is taking the place of Cu2+ in the

solution and kicking it out (displacing it). So the Cu2+ will have no choice but to return to the

solid state Cu(s)0.

We say that Zinc displaced Copper (kicked it out) from its salt solution (We say salt solution

because Cu2+ is not just floating around the solution as an ion, it is actually bound, in the form

of CuSO4). And we say that Zinc was able to kick out Cu2+ because it is a more active metal

than Copper (Recall metals activity series from Year 11), so it wins the battle of being in the

salt solution form (ZnSO4).

Zn(s) + CuSO4(aq) ZnS04(aq) + Cu(s)



What is a displacement reaction?

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For the following chemical reaction:

iron + copper(II) sulfate iron sulfate + copper

a. Write the chemical equation

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b. Is this a displacement reaction? If yes, which metal is being displaced?

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Why does the following reaction not happen?

iron + magnesium sulfate no reaction

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Could you keep a solution of copper sulfate in an iron tank? And Why?

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B. identify the relationship between displacement of metal ions in solution by other metals to the

relative activity of metals

Active metals displace less active metals from its salt solution.

The greater the difference in activity between the two metals, the more vigorous the

displacement reaction

As mentioned in the example of Zinc and Copper earlier, zinc is said to be more active that

copper, therefore it is more likely to react in the solution and form an aqueous solution.

Below is an illustration of the activity series of metals:

http://archives.jesuitnola.org/upload/clark/Refs/act_series.gif

So why is this Activity series useful when it comes to displacement (redox) reactions?

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http://archives.jesuitnola.org/upload/clark/Refs/act_series.gif



Notice changes in oxidation numbers

C. Account for changes in the oxidation state of species in terms of their loss or gain of electrons

The oxidation state is also called the oxidation number.

What is an oxidation state?

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Notice the difference between X2+ X+2

When the charge is written before the number (-2) it means we are talking about oxidation number,

and when it is written after the number (2-) it means we are talking about the charge.

In most cases the oxidation number is the same as the charge (So we can use them interchangeably).

A change in the oxidation number of compounds indicate the transfer of electrons (Redox

reactions).

Oxidation = Loss of electrons (e-)

Reduction = Gain of electrons (e-)

FOR EXAMPLE:

1. Mg0(S) + O2

0(g)

2 Mg+2 O-2

Magnesium is changing its oxidation number from from 0 to +2 and Oxygen from 0 to

-2.

Does this mean that this reaction is a redox reaction? …………………………….….

The best method of balancing redox equations is the half-equation method. What is an oxidising agent?

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What is a reducing agent?

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More on Oxidation numbers

Determine the oxidation number of the following:

Chemical compound Chemical formula Oxidation number

Phosphorus in phosphorus trichloride

Phosphorus in phosphorus pentoxide

Sulfur in sulfur dioxide

Carbon in carbon monoxide



Write Redox half reactions

D. describe and explain galvanic cells in terms of oxidation/reduction reactions

Galvanic cells are also know sometimes as voltaic cells. They are a type of redox reactions that

happen on their own without the need for a source of energy.

Mark S. Cracolice, Edward I. Peters (2011) Introductory Chemistry: An active learning approach, 4th Edition, Brooks/Cole, Cengage Learning,

Page 587.

Why do we connect the two half cells by a salt bridge?

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Examples of Voltaic (Galvanic) cells include:

Figure 19.7 Nickel-cadmium (ni-cad) batteries.

The fundamental components of these batteries

are nickel oxide, nickel hydroxide, and nickel

metal, cadmium hydroxide and cadmium metal,

and potassium hydroxide.

Figure 19.8 Lithium batteries.

These batteries are typically made

from lithium metal and

manganese(IV) oxide.

Mark S. Cracolice, Edward I. Peters (2011) Introductory Chemistry: An active learning approach, 4th Edition, Brooks/Cole, Cengage Learning,

Page 587.

4



E. outline the construction of galvanic cells and trace the direction of electron flow

Oxidation happen at the ANODE (+ve) – AN OX

Reduction happens at the CATHODE (-ve) – RED CAT

The electrons flow from the Anode (Reducing agent) to the Cathode (Oxidising agent)

Each side of the cell is called half-cell. One half represents the oxidation and one half represents the

reduction reactions (See figure above).

Sketch a Galvanic cell.

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F. Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells

Galvanic cell: chemical energy electrical energy

Electrode: part where electrons flow into/out of half-cells

Electrolyte: any solution containing ions

What is a cathode?

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What is an anode?

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Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following:

Button cell

Fuel cell

Vanadium redox cell

Lithium cell

Liquid junction photovoltaic device (eg the Gratzel cell)

In terms of:

Chemistry


Impact on society


Suggestion: Make a table as follows

Cell 1 (eg. Lithium) Cell 2 (Lead-acid cell) Cell 3

Chemistry


Impact on Society




Remember that that all single displacement reactions are redox reactions.

Recall polyatomic ions.

What do half reactions represent?

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Let us take the following single displacement (redox) reaction:

Cu(s) + 2AgNO3 Cu(NO3)2 + 2Ag(s)

Step 1: Assign oxidation numbers

We do that to find out what is being oxidised and what is being reduced in order to

assign the oxidation and reduction half reactions.

According to the rules above (refer to assigning oxidation numbers above), we find that

the oxidation numbers are as follows:

Cu(s)0 + 2Ag+1NO3

-1 Cu+2(NO3-1)2 + 2Ag(s)

0

Step2: Notice the changes in oxidation numbers

Cu(s)0 Cu+2………………………..……. More positive, therefore loosing electrons (Oxidation).

Ag+1 Ag(s)

0………………………..……. More negative, therefore loosing electrons (Reduction)

Step 3: Write the oxidation and reduction reactions taking into account the electrons being

transferred.

Cu(s)0 Cu+2 + 2e- ………………………..……. Oxidation

Ag+1 + 1e- Ag(s)

0 ………………………..……. Reduction

Notice that this reaction is not balanced (The oxidation reaction is losing 2 electrons but the

reduction is gaining only 1).

Writing half-equations for redox reactions

WORKED EXAMPLE



Write half reactions for the following redox reactiona:

NH3 + O2 NO2 + H2O

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2Na(s) + Cl2(g) 2NaCl

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TRY IT YOURSELF



Balance Redox reactions

Remember: Redox reactions happen in solutions (Ions floating around).

In addition to balancing the atoms, we must balance the charges.

To balance redox reactions, we use a methods called Half Equation Method. We separate the

equation into two half-equations (Oxidation and Reduction).

1. Balancing redox reactions in neutral solutions

You will notice in these equations that all protons (H+) will cancel out.

Balance the following reaction

Cu(aq) + Fe(s) → Fe3(aq) + Cu(s)

Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:

Cu(aq) + e− → Cu(s)

Fe3(aq) + 3e− → Fe(s)

The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:

Cu(aq) + e− → Cu(s)

Fe(s) → Fe3(aq) + 3e−

Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu+(aq)+e−→Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives:

3 Cu(aq) + 3e → 3Cu(s)

Fe(s) → Fe3(aq) + 3e−

Step 3: Adding the equations give:

3 Cu(aq) + 3e− + Fe(s) → 3Cu(s) + Fe3 (aq) + 3e−

The electrons cancel out and the balanced equation is left.

3 Cu+(aq) + Fe(s) → 3Cu(s) + Fe3+

(aq)

Worked example

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2. Balancing redox reactions in acidic solutions

In acidic we get hydrogen and oxygen atoms and in addition to balancing the charges (like we

do in neutral solutions) we balance the Hydrogen and Oxygen.

Problem-Solving Strategy

The Half-Reaction Method for Balancing Equations for Oxidation–Reduction

Reactions Occurring in Acidic Solution

1. Write separate equations for the oxidation and reduction half-reactions.

2. For each half-reaction:

A. Balance all the elements except hydrogen and oxygen.

B. Balance oxygen using H2O.

C. Balance hydrogen using H.

D. Balance the charge using electrons.

3. If necessary, multiply one or both balanced half-reactions by an integer to equalize the

number of electrons transferred in the two half-reactions.

4. Add the half-reactions, and cancel identical species.

5. Check that the elements and charges are balanced.



Balancing the equation for the reaction between Permanganate and iron(II) ions in acidic

solution:

1. Assign oxidation numbers and write half reactions noticing the oxidation (loss of electrons) and reduction (gain of electrons)

(Steven S. Zumdahl & Susan A. Zumdahl, 2014)

Reduction reaction

Oxidation reaction

Worked example




Potassium dichromate (K2Cr2O7) is a bright orange compound that can be reduced to a blue-violet solution of Cr31 ions. Under certain conditions, K2Cr2O7 reacts with ethanol (C2H5OH) as follows:

Balance this equation using the half-reaction method.

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2. Balancing redox reactions in basic solutions

Problem-Solving Strategy The Half-Reaction Method for Balancing Equations for Oxidation–Reduction

Reactions Occurring in Basic Solution

1. Use the half-reaction method as specified for acidic solutions to obtain the final

balanced equation as if H ions were present.

2. To both sides of the equation obtained above, add a number of OH ions that is

equal to the number of H ions. (We want to eliminate H by forming H2O.)

3. Form H2O on the side containing both H1 and OH ions, and eliminate the number

of H2O molecules that appear on both sides of the equation.

4. Check that elements and charges are balanced.


Balance the following redox reaction in basic solution:

Worked example



Calculate cell potential

Solve problems and analyse information to calculate the potential requirement of named

electrochemical processes using tables of standard potentials and half-equations.

Cell potential is abbreviated with the symbole: E°. The superscript ° indicates standard-

state conditions.

All standard

Calculating E° Potential for electrochemical cells relies heavily on understanding how to write

half reactions for redox reactions, especially when you are not given the half reactions in the

question. Therefore, we are going to learn how to write half equations for redox reactions.

What is cell potential?

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What is standard cell potential?

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If a standard cell potential is E°cell = +0.85 V at 25 °C, is the redox reaction of the cell

spontaneous?

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What are the factors that affect cell potential?

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PRACTICE QUESTIONS

1. A galvanic cell under standard conditions is represented below. (5 marks)

(a) On the diagram, clearly label the anode, the cathode and the direction of electron flow.

(b) Write a balanced net ionic equation for the overall cell reaction.

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(c) Calculate the standard cell potential (E−°).

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(d) Explain any colour changes observed in this cell as the reaction proceeds.

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2. An electrochemical cell is constructed using two half cells. One half cell consists of an inert

platinum electrode and a solution of Fe2+

and Fe3+

. The other half cell consists of a lead electrode

and a solution of Pb2+

. (7 Marks)

Current will flow from one electrode to the other electrode when the cell is completed using a

voltmeter and a salt bridge.

(a) Write relevant half equations and a balanced net ionic equation for the overall cell reaction.

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(b) Calculate the standard cell potential (E°).

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(c) Identify the anode, cathode, metals and ions by labelling the following diagram

(d) Identify an appropriate electrolyte to use in the salt bridge.

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(4) redox reactions

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