Chemistry For Third Secondary → Electrochemistry and Galvanic cells

Table of Contents

Electrochemistry

Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical reactions.
It deals with the transfer of electrons, ions, and molecules in chemical systems, including the conversion of chemical energy to electrical energy and vice versa.
One important aspect of electrochemistry is the study of galvanic cells, which are devices that use a redox reaction to produce an electric current.
A galvanic cell consists of two half-cells, each containing an electrode and an electrolyte solution. The half-cells are connected by a wire and a salt bridge.
During the operation of a galvanic cell, a spontaneous redox reaction takes place in which electrons are transferred from the reducing agent (or the anode) to the oxidizing agent (or the cathode) through the wire.
The flow of electrons creates an electric current that can be used to do work.
The salt bridge helps to maintain charge neutrality by allowing the flow of ions between the half-cells, preventing the build-up of excess charge.
The voltage produced by a galvanic cell depends on the difference in the electrode potentials of the two half-cells, known as the cell potential or electromotive force (EMF).
The higher the difference in electrode potentials, the greater the cell potential and the stronger the electric current.
Galvanic cells have numerous practical applications, including in batteries, fuel cells, and electroplating.

Oxidation reduction reaction

An oxidation-reduction reaction, also known as a redox reaction, is a type of chemical reaction that involves the transfer of electrons between two or more species.

In an oxidation-reduction reaction, one reactant is oxidized, which involves the loss of electrons, while another reactant is reduced, which involves the gain of electrons.

Oxidation: It is the process of losing electrons, which results in an increase in oxidation state.

example: oxidation of zin by losing 2 electrons :

Zn⁰ ⇒ Zn²⁺ + 2 e^–

Reduction: is the process of gaining electrons, which results in a decrease in oxidation state.

Example: reduction of copper ions by gaining 2 electrons :

Cu⁺² + 2 e^– ⇒ Cu⁰

The reducing agent :In a redox reaction, the species that is oxidized is called the reducing agent,

The oxidizing agent: the species that is reduced is called the oxidizing agent.

Experiment: dip a sheet of zin in blue copper sulphate solution

is the reaction between zinc and copper(II) ions to form zinc ions and copper metal. In this reaction, the zinc atoms lose electrons and are oxidized, while the copper(II) ions gain electrons and are reduced. The electrons are transferred through a wire, and the reaction can be harnessed to do work, such as powering a light bulb.

Observation:

1–Red copper metal precipitated on the surface of zinc sheet

2-Zin-metal dissolves in the solution

3-The blue color of copper sulphate solution will decrease to become colorless

4-The dissolution of Zn increases

Conclusion:

-This reaction is spontaneous redox-reaction :

The reaction consists of two half rection:

a)oxidation reaction :

Zn⁰(S) ⇒ Zn²⁺_(aq) +2e^–

Zinc atom (Zn⁰)loses 2 electrons and converts to zinc ion (Zn²⁺)

– Zn²⁺dissolves and defused in the solution

b)Reduction reaction:

Cu²⁺_(aq) +2e^– ⇒ Cu⁰ _(s) ↓

-copper ion (Cu²⁺ ) accepts 2 electron of Zn and converts to copper metal (Cu⁰ ) that deposited on the surface of the Zn sheet

Galvanic cell ( Daniel cell ):

It consists of : consists of 2 half-cell found in isolated container

1-Cathode(positive pole):copper plate (Cu-electrode) dipped in copper sulphate solution in container

2-Anode(negative pole) :Zinc plate(Zn-electrode) dipped in zinc sulphate solution in another container

GR:

-Zinc electrode is called the anode ?

-Zinc half cell is called the anode have cell ?

Because of the accumulation of electrons produced from oxidation process on its surface

3-Salt bridge: connected between the 2 half-cell

4-Electrolyte:The solution in each container

When the 2 electrodes are connected by a metal wire :

The electrons pass through an electric wire between the 2 halve cells
Electric current produced from the spontaneous oxidation-reduction reaction

What happens :

a)at anode: oxidation

Zn⁰(S) ⇒ Zn²⁺_(aq) +2e^–

b)at cathode :reduction

Cu²⁺_(aq) +2e^– ⇒ Cu⁰_(s)

The total reaction :

-The flow of current is stopped when all of ZINC metal dissolves or all the copper ions disappear GR?

because they are precipitated as a copper atoms (Cu⁰)

-Electrons flow in the outer metal wire from the zinc electrode to the copper electrode

How to write the Diagram of Galvanic cell:

Zn⁰/ Zn²⁺(xM) // Cu²⁺_(xM)/ Cu⁰

The symbols of the diagram:

Single bar :/ indicates a junction between the electrode and electrolytic solution of its ions
Double bar:// indicates a junction between the 2 half cells (or salt bridge)
xM : the value of the concentration of the solution

Example1:

1)In galvanic cell 2 reaction take place write the diagram:

Mg⁰_(s) oxidation Mg²⁺_(aq) + 2e^–

Zn²⁺_(aq)+2e^– reduction Zn⁰_(s)

-The diagram :

Mg⁰_(s) / Mg²⁺ // Zn²⁺_(aq) / Zn⁰_(s)

Example2:

write the diagram of the galvanic cell fro the following reaction :

2Li0 _(s)⇒ 2Li⁺_(aq) +2e- (oxidation)

Ni²⁺_(aq) + 2e^–⇒ Ni⁰_(s)(reduction)

2Li0 _(s) / 2Li⁺ (aq) // Ni²⁺_(aq) / Ni⁰_(s)

Example3:

write the cell diagram of galvanic cell in the following reaction:

Ni²⁺_(aq) + Fe⁰_(s) ⇒ Ni⁰_(s) + Fe²⁺_(aq)

illustrating :

-Anode and Cathode

-direction of the flow of the electric current in the outer wire

Answer:

Fe⁰_(s) / Fe²⁺_(aq) // Ni²⁺_(aq) / Ni⁰_(s)

The direction of electric current :from Anode to Cathode

The electric current flows from the anode (Fe) to the Cathode (Ni)

Note:

-The oxidation process occurs at the anode

-The reduction process occurs at the cathode

What is a salt bridge ?

-a glass U-shaped tube filled with strong electrolytic solution like (Na₂SO₄)

Its ions don’t react with the ions present in the 2 half cells and the electrodes materials

Function of salt bridge:

1-it connects between the solution of the 2 half cells indirectly

2-it neutralize the excess of both the positive and negative ions which forms in the solutions of the 2 half cells which prevents their accumulation

3-The absence of the salt bridge lead to stopping the oxidation and reduction reaction and stopping the electric current

What is effect of absence of salt bridge?

-The flow of electric current stop due to the stopping of the spontaneous oxidation reduction reaction as a result of the accumulation of the excessive ions in the solution of the 2 half cells

When would the flow of the electric current stop in Daniell cell?

It stops in the following cases:

1-The full consumption of the zinc electrode and its dissolution in the electrolytic solution

2-The full consumption of the copper(II) cations Cu²⁺ (aq) from the solution of the copper half cell and their deposition on the copper electrode as copper atoms

3-Removing the salt bridge

What would happen on?

replacing sodium sulphate solution in the salt bridge present in Daniell cell with barium chloride solution ?

-The electric current stop because ion of Ba²⁺ combine with sulphate ions (SO4^2-) in the solution of copper half forming barium sulphate precipitate

How to Calculate the electromotive force (emf) of galvanic cells ?

E _cell (emf) = E⁰_oxid (Anode) + E⁰_red (Cathode)

= E⁰_red (Cathode) – E⁰_red (Anode)

= E⁰_oxid (Anode) – E⁰_oxid (Cathode)

E⁰_oxid = – E⁰_red of the same electrode

Example:

Zn²⁺ +2 e^– ⇒ Zn⁰_(s) E _red=-0.76 V

Zn ⁰_(s)⇒ Zn²⁺_(aq) + 2 e ^– E⁰_oxid =+ 0.76

—————————————————————

Cu²⁺_(aq) + 2 e^– ⇒Cu⁰_(s) E⁰_red = +0.34 V

Cu⁰_(s) ⇒ Cu²⁺_(aq) + 2 e^– E⁰_oxid = -0.34 V

Notes_:

-The half cell with the higher oxidation potential represents the anode half cell(The negative electrode at which the oxidation occurs)

-The half cell with the higher reduction potential represents the cathode half cell (the positive electrode at which the reduction occurs)

Calculate the electromotive force of a Daniell cell by knowing :

	oxidation potential	reduction potential
Zinc half cell	+0.76 V	-076 V
Copper half cell	-0.34 V	+0.34 V

Answer:

oxidation potential of zinc is higher so The zinc half cell represents the anode half cell

Reduction potential of copper is higher so the copper half cell is the cathode half cell

E cell(emf)=E⁰_oxid (Anode + E⁰_red (Cathode)

=E⁰_oxid (Zn⁰) + E0 _red (Cu²⁺)

= 0.76+0.34=1.1 V

Another answer:

E _cell(emf)= E⁰ red (athode) – E⁰red (anode)

=E⁰_red (Cu²⁺) – E⁰_red (Zn²⁺)

= 0.34 –(-0.76)=1.1 V

Example2:

A galvanic cell consist of a half cell :

Sn⁰ /Sn²⁺ (aq) and a half cell Ag⁰_(s) / Ag⁺_(aq)

If you know that :

The standard reduction potential of tin ions = -0.14 V

The standard reduction potential of silver ions =+0.8 V

1)write the cell diagram 2) calculate (eme) of this cel

Answer:

1)The reduction potential of silver ion is higher so it is the Cathode half cell and tin is the anode half cell

2Ag⁺ +2 e^– ⇒ 2Ag⁰_(s)

Sn⁰_(s) ⇒ Sn²⁺_(aq) + 2 e^–

The cell diagram is :

Sn⁰ /Sn²⁺_(aq) // 2Ag⁺_(aq) / 2Ag⁰ (s)

2)emf =E⁰_red (Ag⁺) – E⁰ red (Sn²⁺)

=0.8 –(-0.14)=0.94 V

Example3:

A and B are 2 elements both of them are divalent their oxidation potentials are -0.3 v ,+0.7 respectively :

1)write the cell diagram of the galvanic cell which consists of these two elements

2)calculate (emf) of this cell would it produce electric current ?Explain

Answer:

1)The oxidation potential of B element is higher so B is the anode half cell and A is the cathode half cell

B⁰_{(s) ⇒}B²⁺_(aq) + 2 e^–

A²⁺_(aq) +2 e^– ⇒ A⁰_(s)

So the cell diagram is:

B⁰_(s) / B²⁺_(aq) // A²⁺_(aq) / A⁰_(s)

2)emf= E⁰_oxid (B⁰) – E0 _oxid (A⁰)

=0.7 –(-0.3) 1 V

The cell produce an electric current because the value of (emf) is positive which means that this is a galvanic cell (oxidation –reduction is spontaneous)

How to measure the Electrode Potentials?

–The there is no certain and direct way to measure the absolute electric potential difference between metal electrode and solution of its ions

To measure the difference between the two electrodes potential of galvanic cell:

By forming galvanic cell consists of:

1)standards electrode(known potential):as The standard hydrogen electrode (its potential = ZERO volt)

2)Unknown electrode: that required to measure its potential

By measuring the electromotive force of the cell (cell potential) we can calculate the potential of unknown electrode

The standard hydrogen electrode: consists of

1-a platinum sheet(1cm2)

2-covered by a black spongy layer of platinum

3-dipped in a solution of a molar concentration (1mol/l) of strong acid

4-passing a flow of H2 gas at(1atm) over the sheet

5-In these condition the hydrogen electrode is called the standard hydrogen electrode(SHE)

6-Its potential=ZERO volt

When the potential of hydrogen changes?

if the conditions change such as partial pressure of hydrogen changes from (1atm) and concentration of hydrogen ions changes from 1M

– Any element except hydrogen has potential difference between the element and its ions

Diagram of the standard hydrogen half-cell:

Pt-H₂ (1atm) / 2H⁺ (1mol/L)

-On using standard hydrogen electrode as an (anode):

Pt +H₂_(g) (1atm) / 2H⁺_(aq) (1M)

-On using standard hydrogen electrode as a cathode):

2H⁺_(aq) (1M) / Pt + H₂ (g) (1atm)

GR: Scientists use the standard hydrogen electrode (SHE) as a standard electrode ?

–because its standard reduction potential (as well as its standard oxidation potential ) under the standard condition equals ZERO

Example:

If the cell diagram of a galvanic cell is represented as:

2Li _(s) / 2Li₊_(aq) // 2H⁺_(aq) (1M) / pt + H₂_(g) (1 atm)

Define :

1)The cathode in this cell 2) The oxidizing agent in the reaction

3)The oxidation potential of lithium knowing that (emf) of the cell equals 3.04 V

Answer:

1)The standard hydrogen electrode

2)The positive hydrogen ions

3)emf = E⁰_oxid (Li⁰) + E⁰^red (H⁺)

3.04 = E⁰_oxid (Li⁰) + Zero so E⁰_oxid (Li⁰)=3.04 V

The Electromotive series

The arrangement of the elements descndingly relative to the negative reduction potentials and ascending relative to the positive reduction potentials whereas the higher negative valued lie at the top of the series and the higher positive values lie at the bottom of it

The arrangement of the elements descendingly relative to the positive oxidation potentials and ascending relative to the negative oxidation potentials whereas the higher positive values lie at the top of the series and the higher negative values lie at the bottom of it .

Elements lie above

Elements lie down

1)have negative reduction potential

2)have positive oxidation potentials

3)Easily to be oxidized GR?

-because they lose their valence electrons easily when react with ions of element in lower position in the series

4-strong reducing agent

5-strength of the reducing agent increases by increasing its oxidation potentials

1)have positive reduction potential

2)have negative oxidation potentials

3)Easily to be reduced GR?

-because they have more ability to accept electrons when react with ions of element in higher position in the series

4-strong oxidizing agent

5-strength of the oxidizing agent increases by increasing its reduction potentials

i⁰_(s) ⇒ Li⁺ + e^–

E⁰_oxid =+3.045 V

Na⁰_(s) ⇒ Na⁺_(aq) + e^–

E⁰_oxid =+2.714

Lithium is a stronger reducing agent than sodium because its oxidation potential is higher

F⁰₂_(g) +2e^– ⇒ 2F^–_(aq)

E⁰_red = +2.87 V

Cl₂⁰_(g) + 2 e^– ⇒2 Cl^–_(aq)

E⁰_red = +1.36 V

-Fluorine is a stronger oxidizing agent than chlorine because its reduction potential is higher

The relation between the strength of the oxidizing agent and its reduction potential and the strength of reducing agent and its oxidation potential

GR:

1-It is easy to reduce fluorine gas F₂?

-because its reduction potential (+2.87V) is the highest compared to other elements so it’s the strongest oxidizing agent

2- It is easy to oxidize lithium element Li (s) ?

-because its oxidation potential (+3.04 V) is the highest compared with other elements so it is the strongest reducing agent

-The elements at the top of the series replace the elements follow them in their salt solution:

Zn _(s) + CuSO4 _(aq) ⇒ ZnSO4 _(aq) + Cu _(s)

Fe _(s) + CuSO4 _(aq) ⇒ FeSO4 _(aq) + Cu _(s)

Cu _(s) + FeSO4 _(aq) ⇒ No reaction GR?

-The ability of the element to replace the following element increases by increasing the distance between them

So The ability of Zn to replace Cu is more than that of Fe

-The elements above hydrogen can replace hydrogen ions in acid or water

Fe _(s) + 2HCl _(aq) ⇒ FeCl _(aq) + H₂_(g)

complete: Cu + HCl ⇒ …………..

GR: Cu can not replace hydrogen of water or diluted acids while sodium replaces hydrogen of water and acids

Mention the importance of the electromotive series?

1-determination of the possibility of the occurrence of the oxidation –reduction between 2 elements

2-determination the ability of elements to replace each other

3-Determination of (emf) value of galvanic cell

Example: arrange the following half cells ascending according to their strength as reducing agents :

* Zn ²⁺_(aq) / Zn⁰_(s) E⁰=-0.76 V * Mg²⁺_(aq) / Mg⁰_(s) E⁰ = -2.37 V

* 2Cl^–_(aq) / Cl⁰²_(g) E⁰=-1.36V * K⁺_(aq) / K⁰_(s) E⁰ = -2.925 V

* Pt²⁺_(aq) / Pt⁰_(s) E⁰ = +1.2 V

Then write the cell diagram of the galvanic cell producing the highest electromotive force among the previous half cells showing the direction of the flow of the electric current .

Answer: Strength of the reducing agent increases by increasing its oxidation potential So: the half cell can arranged asecndingly according to their strength as reducing agent :

-The galvanic cell that produce the highest E.m.f is the potassium/chlorine cell :

E cell = E _oxid (Anode) + E _red (Cathode) = 2.925+1.36=4.285 V

The cell diagram is K⁰_(s)/ K⁺_(aq) // Cl⁰₂_(g) / 2Cl^–_(aq)

– The electric current flows from the potassium (anode) to the Chlorine(cathode)

Quiz about galvanic cell

A galvanic cell is a device that uses a spontaneous redox reaction to generate an electric current.
Another name for a galvanic cell is a voltaic cell.
The purpose of a galvanic cell is to convert chemical energy into electrical energy.
A galvanic cell generates electricity by converting the energy released in a spontaneous redox reaction into an electrical potential difference.
An anode in a galvanic cell is the electrode where oxidation occurs.
A cathode in a galvanic cell is the electrode where reduction occurs.
An electrolyte in a galvanic cell is a solution containing ions that can conduct electricity.
A salt bridge in a galvanic cell is a device that connects the two half-cells and allows ions to flow between them.
The function of a salt bridge in a galvanic cell is to maintain charge neutrality in the half-cells and to prevent the mixing of the two electrolytes.
The main difference between a galvanic cell and an electrolytic cell is the source of electrical energy. In a galvanic cell, the electrical energy is generated by the redox reaction, while in an electrolytic cell, the electrical energy is supplied externally to drive a non-spontaneous redox reaction.
A primary galvanic cell is a cell that can only be used once, while a secondary galvanic cell is a cell that can be recharged and used multiple times.
The Nernst equation is a mathematical equation that relates the concentration of reactants and products in an electrochemical cell to the cell potential.
The standard reduction potential is the potential of a half-reaction relative to the standard hydrogen electrode (SHE) under standard conditions (1 M concentration and 1 atm pressure).
The standard hydrogen electrode is a reference electrode that has a potential of 0 V under standard conditions and is used as a reference to measure the potential of other electrodes.
The standard cell potential can be calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.
The relationship between cell potential and Gibbs free energy is given by the equation ΔG = -nFE, where ΔG is the change in Gibbs free energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the cell potential.
Faraday’s constant is the amount of charge carried by one mole of electrons, and its value is 96,485 C/mol.
The relationship between charge and Faraday’s constant is given by the equation Q = nF, where Q is the charge, n is the number of moles of electrons, and F is Faraday’s constant.
Corrosion is the process of degradation of a material due to a chemical reaction with its environment. Galvanic cells can cause corrosion when two dissimilar metals are in contact with each other in the presence of an electrolyte.
Galvanic cells are used in batteries to provide a source of electrical energy. Batteries contain one or more galvanic cells that convert chemical energy into electrical energy to power devices such as phones, laptops, and cars.

Answers of quiz

A galvanic cell is a device that uses a spontaneous redox reaction to generate an electric current.
Another name for a galvanic cell is a voltaic cell.
The purpose of a galvanic cell is to convert chemical energy into electrical energy.
A galvanic cell generates electricity by converting the energy released in a spontaneous redox reaction into an electrical potential difference.
An anode in a galvanic cell is the electrode where oxidation occurs.
A cathode in a galvanic cell is the electrode where reduction occurs.
An electrolyte in a galvanic cell is a solution containing ions that can conduct electricity.
A salt bridge in a galvanic cell is a device that connects the two half-cells and allows ions to flow between them.
The function of a salt bridge in a galvanic cell is to maintain charge neutrality in the half-cells and to prevent the mixing of the two electrolytes.
The main difference between a galvanic cell and an electrolytic cell is the source of electrical energy. In a galvanic cell, the electrical energy is generated by the redox reaction, while in an electrolytic cell, the electrical energy is supplied externally to drive a non-spontaneous redox reaction.
A primary galvanic cell is a cell that can only be used once, while a secondary galvanic cell is a cell that can be recharged and used multiple times.
The Nernst equation is a mathematical equation that relates the concentration of reactants and products in an electrochemical cell to the cell potential.
The standard reduction potential is the potential of a half-reaction relative to the standard hydrogen electrode (SHE) under standard conditions (1 M concentration and 1 atm pressure).
The standard hydrogen electrode is a reference electrode that has a potential of 0 V under standard conditions and is used as a reference to measure the potential of other electrodes.
The standard cell potential can be calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode.
The relationship between cell potential and Gibbs free energy is given by the equation ΔG = -nFE, where ΔG is the change in Gibbs free energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the cell potential.
Faraday’s constant is the amount of charge carried by one mole of electrons, and its value is 96,485 C/mol.
The relationship between charge and Faraday’s constant is given by the equation Q = nF, where Q is the charge, n is the number of moles of electrons, and F is Faraday’s constant.
Corrosion is the process of degradation of a material due to a chemical reaction with its environment. Galvanic cells can cause corrosion when two dissimilar metals are in contact with each other in the presence of an electrolyte.
Galvanic cells are used in batteries to provide a source of electrical energy. Batteries contain one or more galvanic cells that convert chemical energy into electrical energy to power devices such as phones, laptops, and cars.