Notes Chem New

UNIT-3 Electrochemistry

Electrochemistry is the study of production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to bring about non-spontaneous chemical transformations.

Electrochemical Cells OR Galvanic Cells OR voltaic cell A galvanic cell is a device that converts the chemical energy of a spontaneous redox reaction into electrical energy.

Uses: 1. In this device the Gibbs energy of the spontaneous redox reaction is converted into

electrical work which may be used for running a motor or other electrical gadgets like heater, fan, geyser, etc.

2. Electrochemical cells are extensively used for determining the pH of solutions, solubility product, equilibrium constant and other thermodynamic properties and for potentiometric titrations

Example of galvanic cell :Daniel cell

1. Reaction: Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)

2. Electrical potential = to 1.1 V ( when concentration of Zn2+and Cu2+ ions is unity).

3. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution.

4. The other half-cell in which reduction takes place is called cathode and it has a positive potential with respect to the solution.

5. These two portions of the cell are also called half-cells or redox couples

(i) Cu2+ + 2e - Cu(s) (reduction half reaction)

(ii) Zn(s) Zn2+ + 2e – (oxidation half reaction)

The net reaction is the sum of two half cell reactions.

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) 6. Electrons flow from anode to cathode in the

external circuit.

7. Inner circuit is completed by the flow of ions through the salt bridge.Charge carriers in Inner circuit - ions and in the external circuit- electrons

2 | P a g e I L I A S t u t o r i a l

Construction of Galvanic cell: 1. Each half cell consists of a metallic electrode dipped into an electrolyte.It is called

electrode

2. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally.

3. The electrolytes of the two half-cells are connected internally through a salt bridge

4. Sometimes, both the electrodes dip in the same electrolyte solution and in such cases we don’t require a salt bridge.

Function of Salt Bridge: Salt bridge (inverted U tube contains an inert electrolytic solution -KCl, KNO3, NH4NO3)1. To maintain electrical neutrality of the solution in the two half cell.2. To complete the electrical circuit by allowing the ions to flow from one solution to the other with out mixing of the two solutions.

Representation of Galvanic cell: -----------------------------------------------------------------------------

Left Salt Bridge Right

Anode Cathode

Oxidation Reduction

Loss of electrons Gaining of electrons

Negative Positive

M (S) | Mn+ (conc.) | | Mn+ (conc.)

| M(S)

Q1. Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq) . Zn2+(aq)+2Ag(s) takes place. Further show:(i) Which of the electrode is negatively charged?(ii) The carriers of the current in the cell.(iii) Individual reaction at each electrode.

Ans: (1) Zn is being oxidised , therefore it acts as anode , hence it is negatively charged.(2) Zn+2 and Ag+ (3) Two half cell reactions are

Zn(s) Zn+2 (aq) + 2e 2Ag+ (Aq) + 2e Ag(s)

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Electrode potential(E): The potential difference developed between the electrode and the electrolyte is called electrode potential.M+n + ne → M(s) EM

+n/M(S) for metallic electrode

X2 + ne → 2X- EX2/ X- for non-metal electrode

Standard electrode potential(E O ) : The electrode potential of a metal electrode as determined with respective to a standard or normal hydrogen electrode when the concentrations of all the species involved is unity.NOTES :

According to IUPAC rules Reduction potentials is taken Standard electrode potential(E O )

Oxidation potential and reduction potent numerical value is equal but algebraic sign is different

E0 value does not change with react reaction quotient changes with change in concentration and temperature

Electromotive force (emf) : The potential difference between the two electrodes of a galvanic cell is called the cell potential when no current is drawn through the cell.

Cell Potential: The potential difference between the two electrodes of a galvanic cell is called the cell

potential (Or) the difference between the electrode potentials (reduction potential) of the cathode and anode.

It is measured in volts.

Eo cell = Eo right – Eo left

Question: Write the half cell reactions and depict the cell for the cell reaction: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2 Ag(s)

Half-cell reactions: Cathode (reduction): 2Ag+ (aq) + 2e– 2Ag(s) Anode (oxidation): Cu(s) Cu2+ (aq) + 2e– Cu(s) |Cu2+ (aq) ||Ag+ (aq) |Ag(s)

Ecell = Eright – Eleft = E (Ag+/Ag) – E (Cu2+/Cu)


Eo cell = Eo (cathode) – Eo (anode)


Measurement of Electrode Potential: The potential of individual half-cell cannot be measured. We can measure only the difference between the two half-cell potentials that gives the

emf of the cell. If we arbitrarily choose the potential of one electrode (half- cell) then, that of the other

can be determined with respect to this. For this purpose we use, a half-cell called standard hydrogen electrode, represented

by Pt(s)H2(g)/H+(aq). It is assigned a zero potential at all temperatures

corresponding to the reaction H+ (aq) + e– ½ H2 (g)

Construction of SHE / NHE : The standard hydrogen electrode consists of a platinum electrode

coated with platinum black. The electrode is dipped in an acidic solution and pure hydrogen gas

is bubbled through it. The concentration of both the reduced and oxidised forms of hydrogen

is maintained at unity. (This implies that the pressure hydrogen gas is one bar and the concentration of hydrogen ion in the solution is one molar.)

Measurement of electrode potential of Zn 2+ / Zn Electrode: Construct a cell At 298 K and find the emf of the cell,

Standard hydrogen electrode | | Zn2+ / Zn is constructed by taking standard

hydrogen electrode as anode and Zn2+ / Zn as cathode, gives the reduction potential of this half-cell.

If the concentrations of the oxidised and the reduced forms of Zn2+ / Zn, the species

in the right hand half-cell are unity, then the cell potential is equal to standard electrode potential,

of the given half-cell.

As for standard hydrogen electrode is zero.

The measured emf of the cell: Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) | | Zn2+ (aq,

1M) | Zn is -0.76V



Measurement of electrode potential of Cu 2+ / Cu Electrode:

SHE is connected at the left (anode) and Cu2+/ Cu electrode at the right (cathode) gives

the reduction potential of this half-cell.

If the concentrations of the oxidized and the reduced forms of Cu2+ / Cu the species in

the right hand half-cell are unity, then the cell potential is equal to standard electrode

potential of the given half-cell.

As for standard hydrogen electrode is zero.

The measured emf of the cell: Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) | |Cu2+ (aq, 1 M) | Cu

is 0.34 V

Different types of electrodes:Sometimes metals like platinum or gold are used as inert electrodes. They do not participate in the reaction but provide their surface for oxidation or reduction reactions and for the conduction of e electrons.

For example, Pt is used in the following half-cells: Hydrogen electrode: Pt(s)| H2 (g) | H+ (aq) With half-cell reaction

H+ (aq) + e– ½ H2 (g)

Bromine electrode: Pt(s) | Br2 (aq) | Br–(aq) With half-cell reaction:

½ Br2 (aq) + e– Br–(aq)

Electrochemical Series:The arrangement of elements in order of increasing reduction potential values is known as electrochemical series(highest negative EO value placed at the top and highest positive E O value is placed at the bottom)

Applications of electrochemical series:1. Relative strengths of oxidizing and reducing agents:

Substances which have lower reduction potential (- ve E º) are stronger reducing agents (at the bottom of the series) eg. Li is the strongest reducing agent. (Reacts by lose of electrons-have least tendency to get reduced)



Substances which have higher reduction potential (+ve E º) are stronger oxidizing agents (at the top of the series) eg. F2 is the strongest oxidizing agent. (Reacts by gaining of electrons-have maximum tendency to get reduced).

2.Comparison of reactivities of metals:A metal having smaller reduction potential (lower position) can displace metals having larger reduction potential from the solution of their salts.

Q. Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.(PAGE NO:

Ans: Any metal will be able to displace any other metal which is placed below it in the electrochemical series.Mg will displace all metalsZn will displace all metals except MgFe will displace Cu and AuCu will displace only AuAu will displace none.

Q. Knowing that :

Cu2+ (aq) + 2 e– ——— Cu (s) E° = + 0.34 V

2 Ag+ (aq) + 2 e– ——— 2 Ag (s) E° = + 0.80 V

Reason out whether, 1 M AgNO3 solution can be stored in Copper Vessel or 1 M CuSO4 solution in Silver Vessel.

3. Calculation of the E.M.F.of the cell : Steps:

1. Write the two half cell reaction2. Equalize the number of electrons in the two equations by multiplying the

equation by a suitable number. (E0 are not multiplied)


Q. Given the standard electrode potentials,K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V. Arrange these metals in their increasing order of reducing power.

Ans:Ag<Hg,Cr,Mg<,KPotassium is the strongest reducing agent while silver is the weakest reducing agent. Higher the oxidation potential, more easily it is oxidized, hence Greater is the reducing power.


3. Use the formula :4. Predicting feasibility/ spontaneity of a redox reaction:

1. In general a redox reaction is feasible only if the species which has higher reduction potential is reduced .i.e. accepts the electron and the species which has lower reduction potential is oxidized. i.e. loses the electron.

2. The emf of the cell based upon the given redox reaction is calculated.

3. If emf comes out to be negative the direction as given, cannot take place but the reverse reaction may take place spontaneously.

Using the standard electrode potentials given in Table, predict if the Reaction between the following is feasible:

EOFe+3/Fe+2 EOI2/I- EOAg+/Ag EOCu+2/Cu EO Br2/Br- 0.77v 0.54v 0.80v 0.34v 1.o9v

(i) Fe3+(aq) and I–(aq)(ii) Ag+ (aq) and Cu(s)(iii) Fe3+ (aq) and Br– (aq)(iv) Ag(s) and Fe 3+ (aq)(v) Br2 (aq) and Fe2+ (aq).

5. Predicting whether a metal can liberate hydrogen from acid or not:In order that a metal M may react with an acid to give H2 gas, the following reactions should take place:

M(s) + H+-----M+ + ½ H2

M--- M+ +e-(oxidation)

H+ +e----½ H2 (reduction)

The metal should have the tendency to lose electrons (negative reduction potential-undergo oxidation), with respect to hydrogen.

Note: Cu does not dissolve in HCl. In Nitric acid it is oxidized by nitrate ion and not by hydrogen ion.

1 How would you determine the standard electrode potential of the system Mg2+/Mg?

2 Can you store copper sulphate solutions in a zinc pot?

3 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. EOFe+2/Fe+3 = -0.77V

4 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.



5Given the standard electrode potentials,

K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V. Arrange these metals in their increasing order of reducing power.

6

Depict the galvanic cell in which the reaction

Zn(s)+2Ag+(aq) . Zn2+(aq)+2Ag(s) takes place. Further show:

(i) Which of the electrode is negatively charged?

(ii) The carriers of the current in the cell.

(iii) Individual reaction at each electrode.7

Knowing that :

Cu2+ (aq) + 2 e– ——— Cu (s) E° = + 0.34 V

2 Ag+ (aq) + 2 e– ——— 2 Ag (s) E° = + 0.80 V


8Can a nickel spatula be used to stir a solution of Copper Sulphate ? Justify your answer.

(E°Ni²+/Ni = – 0.25 V E°Cu²+/Cu = 0.34 V)

9.

11

12

13



Reversibility of cell by applying external opposite potential to Galvanic cell: If an external opposite potential is applied and increased slowly, we find that the reaction

continues to take place When the opposing voltage reaches the value 1.1 V then, the reaction stops altogether and

no current flows through the cell. Further increase in the external potential again starts the reaction but in the opposite

direction. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions.

1Consider the electrochemical cell :

Zn (s) / Zn2+ (aq) // Cu2+ (aq) / Cu. It has an electrical potential of 1.1 V when

concentration of Zn2+ and Cu2+ ions is unity.State the direction of flow of electrones and also specify if Zinc and Copper are

deposited or dissolved at their respective electrodes. When :

(a) an external opposite potential of 0.8 V is applied.(b) an external opposite potential of 1.1 V is applied



NERNST EQUATION

The standard electrode potentials (given in table) are measured when the concentration of all the species involved in the electrode reaction is unity and temperature is 298K.

This need not be always true. Nernst showed that the electrode potential depends on the concentration of the electrolyte solution and temperature.

NERNST EQUATION FOR SINGLE ELECTRODE

For the electrode reaction:M n+ (aq) + ne– M(s)

The electrode potential at any concentration measured with respect to standard hydrogen electrode can be: represented by since [M(s)] =1

(R is gas constant (8.314 JK–1 mol–1) F is Faraday constant (96,487 C mol–1), T is temperature in

kelvin )and [Mn+] is the concentration of the species, Mn+.

1. a) For the reaction Zn2+ (aq) +2e----- Zn(s) Write the Nernst equation b) For the reaction Cu+2 + 2e →Cu write the single electrode potential

NERNST EQUATION FOR CELL

In Daniell cell, the electrode potential for any given concentration of Cu2+and Zn2+ ions,Zn(s)+2Cu+2(aq) → Zn2+(aq)+2Cu(s) we write For cathode:

Converting the natural logarithm to the base 10 and substituting R, F and T=298K



We should use the same number of electrons (n) for both the electrodes.

It can be seen that E (cell) depends on the concentration of both Cu2+and Zn2+ions. It increases with increase in the concentration of Cu2+ions and decrease in the concentration of Zn2+ions.

1. Write the Nernst equation for following cell

Solution

2.

solution



Equilibrium Constant from Nernst Equation: If the circuit in Daniell cell is closed then we note that the reaction: Zn(s) + Cu2+ (aq) ----Zn2+ (aq) + Cu(s) takes place and as time passes

the concentration of Zn2+ keeps on increasing while the concentration of Cu2+ keeps on decreasing.

The voltage of the cell as read on the voltmeter keeps on decreasing. After some time, there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. (it means Ecell at this concentration zero)

This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as

1.

Electrochemical Cell and Gibbs Energy of the Reaction:

Gibbs Energy of the Reaction : It is the measure of the maximum useful electrical work that can be obtained from a chemical reaction.

As reversible work is done by a galvanic cell it results in decrease in its Gibbs energy.

Electrical work done = Decrease in Gibbs energy Electrical work done in one second = Electrical potential X Total charge

passed. If the emf of the cell is E and nF is the amount of charge passed and ΔrG is

the Gibbs energy of the reaction, then



E (cell) is an intensive parameter (independent on the amount of substance) but ΔrG is an extensive thermodynamic property and the value depends on n.

For the reaction Zn(s) + Cu2+ (aq) --Zn2+ (aq) + Cu(s) ΔrG = -2FE (cell)

But when we write the reaction 2 Zn (s) + 2Cu2+ (aq) ---2 Zn2+ (aq) +2Cu(s) ΔrG = –4FE (cell)

But when we write the reaction 2 Zn (s) + 2Cu2+ (aq) ---2 Zn2+ (aq) +2Cu(s)

If the concentration of all the reacting species is unity, then E (cell) = and we have

Thus, from we can obtain the thermodynamic quantity, ΔrGº, standard Gibbs energy of the reaction.

From the latter we can calculate equilibrium constant by the equation:ΔrG º = – RT ln K.

ΔrG º = – 2.303 RT log Kc

1.

a) Calculate the emf of the cell in which the following reaction takes place:Ni(s) + 2Ag+ (0.002 M) Ni 2+

(0.160 M) + 2Ag(s)Given that Eº (cell) = 1.05 V

b) The cell in which the following reaction occurs:

at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

ELECTRO CHEMISTRYCLASS:12( CBSE) ASSIGNMENT :

TOPIC: GALVANIC CELL

1. Depict the galvanic cell in which the reaction From : MOHAMMED.ILIAS visit :meschemistryforummotion.com phone no: 5721312


Cu(s)+2Ag+(aq) . Cu2+(aq)+2Ag(s) takes place. Further show:

1. Write Individual reaction at each electrode2. Which of the electrode is negatively charged?3. The carriers of the current in the cell.

2. Define following terms1. Electrode potential2. Standard electrode potential3. Emf4. Cell potential5. SHE6. Electro chemical series

3 How would you determine the standard electrode potential of the system Mg2+/Mg?4 Can you store copper sulphate solutions in a zinc pot?

5 Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions. EOFe+2/Fe+3 = -0.77V

6 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

7Given the standard electrode potentials,K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V. Arrange these metals in their increasing order of reducing power.

8

9

Knowing that :

Cu2+ (aq) + 2 e– ——— Cu (s) E° = + 0.34 V

2 Ag+ (aq) + 2 e– ——— 2 Ag (s) E° = + 0.80 V


10Can a nickel spatula be used to stir a solution of Copper Sulphate ? Justify your answer.



(E°Ni²+/Ni = – 0.25 V E°Cu²+/Cu = 0.34 V)

11

Using the standard electrode potentials given in Table, predict if the Reaction between the following is feasible:

EOFe+3/Fe+2 EOI2/I- EOAg+/Ag EOCu+2/Cu EO Br2/Br- 0.77v 0.54v 0.80v 0.34v 1.o9v

(i) Fe3+(aq) and I–(aq)(ii) Ag+ (aq) and Cu(s)(iii) Fe3+ (aq) and Br– (aq)(iv) Ag(s) and Fe 3+ (aq)(v) Br2 (aq) and Fe2+ (aq).

12

Consider the electrochemical cell :

Zn (s) / Zn2+ (aq) // Cu2+ (aq) / Cu. It has an electrical potential of 1.1 V when

concentration of Zn2+ and Cu2+ ions is unity.State the direction of flow of electrones and also specify if Zinc and Copper are

deposited or dissolved at their respective electrodes. When :

(a) an external opposite potential of 0.8 V is applied.(b) an external opposite potential of 1.1 V is applied

13

14The standard electrode potential for the Zn+2/Zn is –o.76v .Write the reactions occurring at

Electrodes when coupled with standard hydrogen electrode (SHE)

NERNST EQUATION:

EOFe

+3/Fe+2 EO

I2/I- EO

Ag+

/Ag EOCu

+2/Cu EO Br2/Br

-

0.77v 0.54v 0.80v 0.34v 1.o9vELECTRODES EO

Mg2+ /Mg(s) -2.37vCu2+/Cu 0.34vFe2+/ Fe(s) -0.44Sn2+/ Sn(s) -0.40v



Br2(l)/Br–Cr3+/Cr 0.74Cd2+/Cd -0.40Zn+2/Zn -0.76vAg+ /Ag 0.80v

1 a. Find the zinc electrode potential when Zn+2 concentration is 0.1latM

b. Calculate the single electrode potential of Pt,Cl2/Cl-(0.01m)

2 Write the Nernst equation and emf of the following cells at 298 K:(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.0001 M)|Cu(s)(ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar)| Pt(s)(iii) Sn(s)|Sn2+(0.050 M)||H+(0.020 M)|H2(g) (1 bar)|Pt(s)(iv) Pt(s)|Br2(l)|Br–(0.010 M)||H+(0.030 M)| H2(g) (1 bar)|Pt(s).

3 Calculate the standard cell potentials of galvanic cell in which the followingReactions take place:

(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd

(ii) Fe2+(aq) + Ag+(aq) →Fe3+(aq) + Ag(s)Calculate the . ▲rG and equilibrium constant of the reactions

4 Calculate the potential of hydrogen electrode in contact with a solution whose pH is 105 Calculate the emf of the cell in which the following reaction takes place:

Ni(s) + 2Ag+ (0.002 M) Ni 2+ (0.160 M) + 2Ag(s) →→

Given that Eº (cell) = 1.05 V

6 The cell in which the following reaction occurs:

at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.


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