Unit 10: Electrochemistry — Long Questions

Ans. Electrochemistry

Definition The branch of chemistry which deals with the interconversion of electrical and chemical energy is called electrochemistry

OXIDATION, REDUCTION, AND REDOX REACTIONS

Oxidation

Oxidation is a process involving loss of electron or electrons. It is also the gain of oxygen and loss of hydrogen. But, here oxidation in terms of loss of electrons is discussed.

Examples
Fe+2 —> Fe+3 + e-
Zn0 —> Zn+2 + e-

Here Fe+2 and Zn0 have lost electrons, therefore they are oxidized.

Reduction

Reduction is a process involving gain of electron or electrons. It is also the gain of hydrogen and loss of oxygen. But, here reduction in terms of gain of electrons is discussed.

Examples Cl0 + e- —> Cl-1
Cu+2 + 2e- —> Cu0

In these examples, Cl0 and Cu+2 have gained electrons; therefore, they are reduced.

Oxidation and reduction always take place together. The reaction in which this happens is called redox reactions.

Redox Reaction The reaction in which oxidation and reduction process takes place are called redox reactions.

Example The example of redox reaction is given below.
Zn0 + Cu+2 —> Cu0 + Zn+2

In this reaction zinc undergoes oxidation as it lose two electrons as it goes to zero oxidation state from +2. Copper undergo reduction as it gets two electron and move from +2 to zero oxidation state.

Identify oxidation and reduction in a chemical reaction

There are two ways of finding out whether or not a substance has been oxidized or reduced during a chemical reaction:

• Electron transfer
• Changes in oxidation number

Ans. Oxidation number and its significance

Oxidation number

Definition It is the apparent charge on an atom (per atom) of an element in a molecule or an ion.

• The oxidation number is also called oxidation state.
• Oxidation number is the measure of gain or lose electrons by an element.

Significance of Oxidation number

• An oxidation number is a number given to each atom or ion in a compound that show its degree of oxidation.
• Oxidation number can be positive (+), negative (-) and zero (0).
• The '+' or '-' sign must always be included when writing oxidation numbers.
• A higher positive oxidation number means the atom or ion is more oxidized.
• A higher negative oxidation number means the atom or ion is more reduced.

OXIDATION NUMBER RULES

Oxidation number rules

The oxidation number of any atom or ion can be deduced by using oxidation number rules. It is important to note that an oxidation number refers to a single atom in a compound.

Rules for assigning oxidation numbers

(i) Uncombined elements
The oxidation number of any uncombined (free) element is always zero.
Examples: Oxidation Number of S8, H2, Cl2, Na, Mg, Zn (in elemental form) is always zero (0).

(ii) Electronegativity rule
In a compound or ion, the more electronegative element is given a negative oxidation number.
Example: In HF, fluorine (more electronegative) gets -1, hydrogen gets +1.

(iii) Fixed oxidation numbers for specific groups
In compounds, many atoms or ions have fixed oxidation numbers.
Group 1 (alkali metals) show always +1.
Group2 (alkaline earth metals) show always +2.
Group 17 (halogens) show -1 in binary compounds.
Hydrogen usually show +1, but -1 in metal hydrides (NaH).
Oxygen usually show -2, but in peroxides (H2O2) -1 and in F2O oxygen show +2 oxidation number.

(iv) Monoatomic ions
The oxidation number is equal to the charge on the ion in mono-atomic ions.
Examples: Cl- show oxidation number of -1, A/3+ show oxidation number of +3.

(v) Sum of oxidation numbers in compounds
In a neutral compound, the sum of all oxidation numbers or algebraic sum is always zero.
Examples: In HCl molecule H show +1, Cl show -1 and the sum is 0.
In NaCl the Na show +1, Cl show -1. The sum of oxidation numbers is 0.

(vi) Sum of oxidation numbers in ions
In a polyatomic ion, the algebraic sum of oxidation numbers equals the charge of ion.
Examples: In CO32-, the sum of oxidation numbers is -2.
In SO42-, the sum of oxidation numbers is -2.

(vii) Variable oxidation states
For the transition metals and many non-metals, the oxidation numbers is not fixed, because they show variable oxidation numbers. Their oxidation number are always find by calculations.
Example: The oxidation number of Mn in KMnO4 is +7.

Ans. Determining oxidation numbers in compounds and radicals

General rule

The oxidation number of any atom in a compound or radical can be determined by applying the standard rules.

Compounds of a metal with a non-metal

In such compounds, the metal always has a positive oxidation number and the non-metal has a negative oxidation number

Example Sodium sulfide (Na2S)
Na = +1 (fixed for Group 1 metal
2(+1) + S = 0
+2 + S = 0
S = -2
Oxidation number of S = -2

(i) Compounds of a non-metal with a non-metal

• The compound with two different non-metals sign of oxidation number depends on difference of electronegativity.
• The more electronegative element is given a negative oxidation number.
• The less electronegative element is assigned a positive value accordingly.

Ans. Disproportionation reaction

Definition A disproportionation reaction is a chemical reaction where a single substance acts as both the oxidizing and reducing agent, resulting in two different products with different oxidation states.

Examples

Decomposition of Hydrogen Peroxide

Reaction 2H2O2(l) —> 2H2O(l) + O2(g)

Oxidation number of hydrogen atom in H2O2 = -1
Oxidation number of hydrogen in H2O = -2 (Reduction)
Oxidation number of oxygen in H2O2 = -1
Oxidation number of oxygen in H2O = -3 (Cl is oxidized from 0 to +5)
Oxidation number of oxygen in H2O = -2 (oxygen is reduced oxygen is oxidized from d -1 do 0)
This is a classic disproportionation reaction of chlorine.

OXIDIZING AGENT (OXIDANT) AND REDUCING AGENT (REDUCTANT)

Oxidizing agent An oxidizing agent (atom, ion or molecule) is that substance which oxidizes some other substance, and is itself reduced to a lower oxidation state by gaining one or more electrons.

General form
X (Ox. No. = 0) + e- —> X (Lower Ox. No.)

Example Cl + e- —> Cl- (oxidation number = -1) lower oxidation number

• Cl is reduced from 0 to -1
• Cl acts as an oxidizing agent

General form
M (Ox. No. = 0) —> M+ + e- (Higher Ox. No.)

Example
Na —> Na+ + e-

• Na is oxidized from 0 to +1
• Na acts as a reducing agent

Ans. Carry out the following steps for balancing of redox equations by oxidation number method:

(i) Skeleton equation: Write down the skeleton equation of the redox reaction under consideration.

(ii) Change in oxidation number: Identify the elements, which undergo a change in their oxidation number during the reaction.

(iii) Record: Record the oxidation number above the symbols of the element, which have undergone a change in the oxidation number

(iv) Indicate oxidation number by arrows: Indicate the change in oxidation number by arrows joining the atoms on both sides of the equation. It shows number of electrons gained or lost.

(v) Equate the electrons: Equate the increase or decrease in the oxidation number, i.e., electrons gained or lost by multiplying with a suitable digit.

(vi) Inspection method: Balance the rest of the equation by inspection (hit and trial) method.

Ans. Electrolytic Cell

Definition An electrolytic cell is a device that converts electrical energy into chemical energy through a process called electrolysis.

Main components of an electrolytic cell (constructions).

(i) Electrodes

• Made of metal or graphite (carbon).
• Cathode (-): Negatively charged electrode.
• Anode (+): Positively charged electrode.

(ii) Electrolyte: The electrolyte (the compound that is ionized, it is either

• A molten ionic compound or aqueous solution containing free ions.
• Carries electric current via ion movement.

(iii) Power Supply

• Provides direct current (DC) to drive the non-spontaneous reaction.

(iv) Structure of cell actually depends upon the nature of element extracted.

Working When direct current is applied, one electrode become positively charged (anode) and other become negative charge (cathode)

• Positive ions (cations) migrate to the cathode, where they gain electrons (reduction).
• Negative ions (anions) migrate to the anode, where they lose electrons (oxidation).
• The movement of ions results in chemical changes at the electrodes and overall conversion of electrical energy to chemical energy.

Electrode | Charge | Process | Ions move toward | Reaction
Cathode | Negative | Reduction | Positive ions (cations) | Gain of electrons (e-)
Anode | Positive | Oxidation | Negative ions (anions) | Loss of electrons (e-)

Applications of electrolytic cells

• Electrolysis of NaCl produces sodium metal and chlorine gas
• Metal refining (copper purification)
• Electroplating of metals.
• Production of caustic soda (NaOH) and other chemicals
• Extraction of non-metals like chlorine
• Purification of reactive metals

REDOX REACTIONS IN ELECTROLYSIS

Redox reactions in electrolysis

The passage of electric current drive a non-spontaneous reaction in which compound is taken is called electrolysis.

Electrolysis involves redox reactions, the oxidation and reduction.

• Reduction occurs at the cathode (gain of electrons)
• Oxidation occurs at the anode (loss of electrons)

At the Cathode (-) Reduction happens
Cations (positive ions) move to the cathode, they gain electrons (reduction)
Products may be metal atoms or hydrogen gas, if hydrogen gas is formed it bubbles off.

Examples

Copper ions in solution
Cu2+(aq) + 2e- —> Cu(s)

Copper metal is deposited at the cathode.

Hydrogen ions from acidified water
2H+(aq) + 2e- —> H2(g)

Hydrogen gas bubbles off.

At Anode (+) Oxidation happens

Anions (negative ions) move to the anode. They lose electrons (oxidation) Products may be non-metal gases or other substances.

Examples

Chloride ions
2Cl-(aq) —> Cl2(g) + 2e-

Chlorine gas is released.

Hydroxide ions (from water)
4OH-(aq) —> O2(g) + 2H2O(l) + 4e-

Oxygen gas is released at the anode.

Electrolysis of molten zinc chloride (ZnCl2): In the redox reaction of electrolysis of ZnCl2, the electrode reactions are

Electrode | Reaction | Type of reaction
Cathode (-) | Zn2+(l) + 2e- —> Zn(s) | Reduction
Anode (+) | 2Cl-(l) —> Cl2g + 2e- | Oxidation

Overall cell reaction
ZnCl2(l) —> Zn(s) + Cl2(g)

The electrons lost at the anode are equal to the electrons gained at the cathode, maintaining charge balance.

Ans. Mass of a substance deposited during electrolysis

The mass of a substance deposited (or liberated) during electrolysis is directly proportional to:

• The time over which constant current flows, and the strength of electric current.
• The number of electrons involved in the electrode reaction.

Combining current and time The total charge passed is calculated using:
Q = I × t

Where

• Q= Charge in coulombs (C)
• I= Current in amperes (A)
• t = Time in seconds (s)

The mass of substance produced at (or removed from) an electrode during electrolysis is proportional to the quantity of electricity (Coulombs) which passes through the electrolyte.

The quantity of electricity is expressed in a unit called Faraday symbol (F). the quantity of electric current carried by the charge on 1 mole of singly charged ions.

• 1 Faraday (F) = 96,500 C/mol

Electrolysis of Silver Nitrate (AgNO3), sliver is deposited at cathode.
Ag+(aq) + e- —> Ag(s)

1 mole of electrons (1 Faraday) deposits 1 mole of silver (107.9g).

• Charge required = 1 × 96,500 C.

Electrolysis of Copper(II) Sulfate (CuSO4), copper is deposited at the cathode.
Cu2+(aq) + 2e- —> Cu(s)

1mol 2moles 1mol
2 moles of electrons (2 Faradsays) are needed to deposit 1 mole of copper (63.5g).

• Charge required = 2 × 96,500 C = 193,000 C.

General Formula (Using Moles and Faraday)
No. of Faradsays=No. of moles of electrons (e-) gained or lost.

AMOUNT OF SUBSTANCE PRODUCED DURING ELECTROLYSIS

The value of F can be used to calculate the mass of substance deposited at an electrode and the volume of gas produced at an electrode.

Ans. Avogadro constant and electrolysis

The Avogadro constant NA is the number of particles (atoms, ions, or molecules) in 1 mole of a substance.

The electrolytic method allows us to determine L by calculating the charge associated with 1 mole of electrons and dividing it by the charge on a single electron.

NA = Charge on 1 mole of electrons / Charge on 1 electron

The charge of an electron can be calculated experimentally

• Charge on 1 mole of electrons = 1 Faraday = F = 96,500 C/mol
• The results show the charge on electron is approximately 1.60×10-19 C

OBJECTIVE

To calculate the charge carried by 1 mole of electrons using a simple electrolytic experiment involving copper(II) sulfate solution and copper electrodes.

Apparatus Required

• Pure copper cathode and copper anode
• Aqueous copper(II) sulfate (CuSO4)
• Power supply
• Variable resistor (to control current)
• Ammeter and stopwatch
• Distilled water and propanone (for cleaning electrodes)
• Electronic balance

Procedure

(i) Weigh the copper cathode and anode before electrolysis.
(ii) Set up the electrolytic cell with CuSO4 solution.
(iii) Use a variable resistor to maintain a constant current (0.2 A).
(iv) Pass the current for a fixed time (40 minutes = 2400 seconds).
(v) After electrolysis: Remove the electrodes, wash with distilled water, then with propanone and dry and reweigh both electrodes.

Observations

• The cathode gains mass (copper is deposited).
• The anode loses mass (copper dissolves into solution as Cu2+).
• Prefer using the mass loss of the anode for accurate results (as copper may not stick perfectly to the cathode).

Calculating charge on an electron

A sample calculation is shown below, using a current of 0.20 A for 34 min.

Mass of anode at start of the experiment = 56.530 g
Mass of anode at end of experiment = 56.394 g
Mass of copper removed from anode = 0.136 g
Quantity of charge passed Q = I × t
= 0.20 × 34 × 60
= 408 C

As the amount of copper removed is 0.13 g
So, 0.13 g of copper requires = 408 C

1 g of copper requires = 408 / 0.136

63.5 g of copper requires = (408 / 0.136) × 63.5
= 190500 C

But the equation for the electrolysis shows that 2 moles of electrons are needed to produce 1 mole of copper:

Cu2+(aq) + 2e- —>Cu(s)

The charge on 1 mole of electrons = 190500 / 2 = 95,250C

If the charge on one electron is 1.60 × 10-19 C,

Charge on an electron = 95250 / 6.022 × 1023 = 1.58×10-19 C

This is in good agreement with the accurate value of 6.02 × 1023 mol-1

Hence Avogadro's number is calculated by electrolytic method.

Ans. Electrode potential

Definition When a metal is dipped into solution of its own ions, the electrical potential is established between metal and metal ions in the solution. It is electrode potential and it indicates the case of oxidation or reduction of substance.

During redox processes, a dynamic equilibrium exists between two chemically related species in different oxidation states.

Example When a copper rod is dipped into solution of its ions.
Cu2+(aq) + 2e- <—> Cu(s)

This equilibrium involves two opposing half-reactions:

Opposing reactions at the electrode

(a) Oxidation (Loss of electrons): Parent atoms converting into metal ions from the electrode.

• Copper metal (Cu) loses electrons to form Cu2+ ions: Cu(s) —> Cu2+(aq) + 2e-

• Electrons remain on the surface of the electrode.

(b) Reduction (gain of electrons):

• Ions acting the electrons from metal electrode and get, deposited as metal atom on cathode surface.
• Cu2+ ions in solution gain electrons to form solid copper: Cu2+(aq) + 2e- —> Cu(s)

Redox equilibrium is established When:

• The rate of oxidation becomes equal to rate of reduction.
• This balance defines the electrode potential of the metal.
• A voltage (potential difference) is established between: The metal electrode and the solution of its ions.
• This is called the electrode potential, and it reflects the ease with which a substance can be oxidized or reduced.

Comparison of metals

(i) Unreactive metals (Copper)

• The redox equilibrium lies to the right: Cu2+(aq) + 2e- —> Cu(s)

Cu2+ ions are easily reduced, they readily accept electrons, if equilibrium is compared with equilibrium setup by other metals.

• Copper has a higher electrode potential.

(ii) Reactive metals

• The redox equilibrium lies to the left: Vanadium further V2+(aq) + 2e- —> V(s)

• V2+ ions are harder to reduce, i.e., they accept electrons less readily.
• Vanadium has a lower electrode potential.

ELECTRICAL DOUBLE LAYER

Absolute electrode potentials

• The absolute electrode potential cannot be measured because of formation of electrical double layer when an element is dipped in a solution of its own ion.
• When a metal is placed in a solution containing its own ions: Zn(s) <—> Zn2+(aq) + 2e-

• Zinc atoms lose electrons, forming Zn2+ ions that enter the solution.
• This process leaves excess electrons on the surface of the metal.
• Zn2+ ions accumulate in the solution near the metal surface, they are attracted back to the surface of metal.
• This creates charge an electrical potential between metal and metal ions in the solution.
• This is called the electrical double layer.

Figure shows the formation of double layer forms between the metal and solution interface.

• The potential cannot be measured directly due to formation of double layer.
• The difference in potential of metal ion system and another system can be measured.
• It is called electrode potential and measured in volts.
• So comparison system is used to measure electrode potential called SHE standard hydrogen electrode.

Ans. Standard hydrogen electrode (SHE)

The SHE is a reference half-cell used to measure the standard electrode potential (E°) of other electrodes. It is assigned a value of 0.00 V by definition.

Components of the SHE

(i) Hydrogen Gas (H2) at 1 atm (101 kPa) pressure.
(ii) Hydrochloric acid or any solution providing H+ ions at 1.00 mol dm-3.
(iii) A platinum electrode coated with platinum black, contact with H2 gas and H+1 ions.

• A finely divided platinum provide a surface for electron exchange to H2 gas and H+1 ions in solution.
• It speedup the establishment of equilibrium.
• It is inert electrode.
• Standard electrode potential (E°) for all half-cells are measured in comparison to this electrode.
• SHE is connected to another half-cell, the reading on voltmeter gives the standard electrode potential for that half-cell.

The Half-equation for the hydrogen electrode can be written.
2H+(aq) + 2e- <—> H2(g)

It explains the hydrogen gas and hydrogen ions have equal to lose and gain electron.

Feature | Description
Gas used | H2 at 1 atm
Electrolyte | 1 M H+ (e.g., HCl)
Electrode | Platinum coated with Pt black
Potential | E° = 0.00 V (by definition)
Function | Reference to measure E° of other cells

Ans. Standard Electrode Potential (E°)

Definition The standard electrode Potential (E°) is the voltage of a half-cell, measured under standard conditions, with a standard hydrogen electrode (SHE), as the other half-cell. It reflects a substance's tendency to gain or lose electrons, to be reduced or oxidized.

Standard Conditions

• The voltage of electrochemical cell depends on concentration, temperature and pressure of the gas.
• To ensure fair comparison, the standard conditions these must be strictly maintained, to measure electrode potential.
• Under these conditions, the electrode potential measured is called standard electrode potential. Its symbol is E° and is spoken as 'E standard'.

Condition | Value
Ion concentration | 1.00 mol dm-3
Temperature | 25°C (298 K)
Gas pressure | 1 atm (101 kPa)
Reference electrode | Standard Hydrogen Electrode (SHE)
Symbol used | E°

Purpose of E°

• Compare the oxidizing/reducing power of different electrodes.
• A more positive E° means a stronger tendency to gain electrons (reduction).
• A more negative E° means a stronger tendency to lose electrons (oxidation).

MEASURING STANDARD ELECTRODE POTENTIALS

Measurement of standard electrode potential (E°)

There are three main types of half-cell whose E° value can be obtained when connected to a standard hydrogen electrode (SHE)

Types of half-cells

(i) Metal / Metal ion half-cell

• A metal electrode (e.g. copper strip) is placed in a solution of its ions (e.g. Cu2+).
• Connected to SHE, the measured voltage is +0.34 V
• This means copper has a greater tendency to gain electrons (be reduced) than hydrogen.

Example Cu2+/Cu(s)

Half-equation Cu2+(aq) + 2e- —> Cu(s)
E° = +0.34 V

(ii) Non-metal / non-metal ion half-cell

• Chlorine gas is bubbled over a platinum electrode in a solution containing Cl- ions.
• The platinum electrode is inert and conducts electrons.

Example Cl2(g)/Cl-(aq)

Half-equation Cl2(g) + 2e- —> 2Cl-(aq)

(iii) Ion / ion half-cell

Example Fe3+/Fe2+

• This uses a platinum electrode in a solution containing both Fe2+ and Fe3+ ions.
• No metal is involved; the platinum acts as a conductor

E° = +1.36 V

Half-equation Fe3+(aq) + e- —> Fe2+(aq)
E° = +0.77 V

Measurement of E° of Cu(s) / Cu(aq)

• Connect the half-cell of Cu+2 / cu to the SHE using a salt bridge to allow ion flow.
• Use a voltmeter to measure the cell potential.
• If the SHE is the negative terminal and the other half-cell is positive, the reading is the standard electrode potential (E°) of the test half-cell.
• For Cu2+/Cu, the measured potential is +0.34 V, meaning copper ions readily gain electrons (are easily reduced).

Copper Half-Cell (Cu2+/Cu)

Reaction at copper half-cell Cu2+(aq) + 2e- <—> Cu(s)
E° = +0.34 V

Reaction at hydrogen half-cell (SHE):
2H+(aq) + 2e- <—> H2(g)
E° = 0.00 V

Conclusion

• Cu+2 ions are more likely to gain electrons than H+1 ions.

Ans. Galvanic (Voltaic) Cell
A cell in which the conversion of chemical energy into electrical energy takes place spontaneously in exothermic redox process with oxidation at anode and reduction at cathode.

Cell representation A galvanic cell is designed to take advantage of spontaneous transfer of electrons.

Cu²⁺ / Cu ∥ Zn²⁺ / Zn cell E° = 1.10 V

Construction Two half-cells are connected together, to have a electrochemical cell (galvanic cell). Voltage of this cell. A Cu²⁺/Cu half-cell connected to a Zn²⁺/Zn half-cell to make a complete galvanic cell in which voltage is measured.

Half-cells are connected together using:

• Wires connecting the metal rods in each half-cell to a high-resistance voltmeter: the electrons flow round this external circuit from the metal with the more negative (or less positive) electrode potential to the metal with the less negative (or more positive) electrode potential.

• A salt bridge to complete the electrical circuit allowing the movement of ions between the two half cells so that ionic balance is maintained a salt bridge does not allow the movement of electrons. It prevents accumulation of charges.

• 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.

Voltage of Cell The voltages for the half cells can be represented by the following half-equation:

Cu²⁺₍aq₎ + 2e⁻ ⇌ Cu₍s₎ E° = +0.34 V
Zn²⁺₍aq₎ + 2e⁻ ⇌ Zn₍s₎ E° = −0.76 V

• Working an electrochemical cell is made by connecting a Cu²⁺/Cu half-cell to a Zn²⁺/Zn half-cell.

The relative values of these voltages tell us that Zn²⁺ ions are more difficult to reduce than Cu²⁺ ions.

At Anode Zinc undergoes oxidation, losing electrons to form zinc ions.
Zn₍s₎ ⟶ Zn²⁺₍aq₎ 2e⁻

At Cathode The electrons flows from anode to cathode. Copper ions gain electrons and are reduced to form copper metal.
Cu²⁺₍aq₎ + 2e⁻ ⟶ Cu₍s₎

Overall Reaction The overall cell reaction can be represented as:
Zn₍s₎ + Cu²⁺₍aq₎ ⟶ Zn²⁺₍aq₎ + Cu₍s₎

The process effectively converts the chemical energy of the redox reaction into electrical energy, which can be connected to perform work.

The voltage generated by this cell is +1.10 V

Equation E°cell = E°cathode + E°anode = 0.34−(−0.76) = +1.10

In a Cu-Zn galvanic cell, zinc (Zn) serves as the anode, and copper ions (Cu²⁺) in solution serve as the cathode.

Other Examples
The E° value for the Ag⁺/Ag half-cell is more positive than for the Zn²⁺/Zn half-cell. So, the Ag⁺/Ag half-cell is the positive pole and the Zn²⁺/Zn half-cell is the negative pole of the cell.

Part | Description
Anode | Zinc rod (oxidation occurs here, negative terminal)
Cathode | Copper rod (reduction occurs here, positive terminal)
Salt Bridge | Filter paper soaked in KNO₃ — allows ion flow to balance charges but not electron flow
External Circuit | Electrons flow from Zn to Cu through a wire, powering a device (e.g., bulb, motor)
Voltmeter | Measures the cell potential (1.10 V in this case)

Ans. Applications of E° values

(i) Direction of electron flow
Direction of flow in wires of external circuits can be deduced by comparing E° values of two half cells that make electrochemical series.

Example Ag₍aq₎ + e⁻ ⇌ Ag₍s₎ E° = +0.80V
Zn²⁺₍aq₎ + 2e⁻ ⇌ Zn₍s₎ E° = −0.76V

Electrons flow from • Zinc (Zn) electrode because it has the more negative E° value (−0.76 V), indicating it loses electrons more readily (oxidation).

Electrons flow to • Silver (Ag⁺) electrode because it has the more positive E° value (+0.80 V), indicating it gains electrons more readily (reduction).

Direction of electron flow (External Circuit): Electron flows from Zn²⁺/Zn (anode) to Ag⁺/Ag (cathode) or from negative pole to positive pole.

Fundamental rule in electrochemistry Electrons always flow from the anode to the cathode, from the more negative E° to the more positive E°.

(ii) Feasibility of a reaction using E° values
Standard electrode potential E° gives us the measure of how easy or difficult it is oxidized or reduced.

E° values are written as reduction half-equations by convention.

For each half equation, more oxidized form is on the right side and more reduced form on the right.

To check the feasibility of reaction between
• The more positive E° value, the greater tendency for the half equation to proceed in forward direction, easier it is to reduce the specie on the left of half equation. The more positive the E° value, the stronger the oxidizing agent.

• The less positive E° values, the greater tendency for the half equation to proceed in reverse direction, easier it is to oxidize the specie on the right of half equation. The less positive (or more negative) the E° value, the stronger the reducing agent.

Reaction Example Mg²⁺ + Ag → ?

We need half-cell equations and E° values
Mg²⁺ + 2e⁻ ⇌ Mg E° = −2.32V
Ag⁺ + e⁻ ⇌ Ag E° = +0.80V

So,
E°cell = E°red − E°oxi
E°cell = (−2.32) − (0.80) = −1.52 V → Not feasible (reaction not spontaneous)

Reaction Example Ag⁺ + Mg → ?
Negative value of E°cell indicates the reaction is not feasible. On the other hand reaction between Ag⁺ and Mg is feasible as for this reaction E°cell will be positive as given below.

E°cell = E°red − E°oxi
−0.80 − (−232)
E°cell = + 1.52 V

The feasible reaction is 2Ag₍aq₎ + Mg₍s₎ ⟶ Mg²⁺₍aq₎ + 2Ag₍s₎

Ans. Variation of E° value with ion concentration

• The position of an equilibrium reaction is affected by changes in concentration, temperature and pressure.

• If we change the concentration or temperature of half-cell X, the electrode potential also changes.

Standard electrode potential (E°) is measured under standard conditions:
• Ion concentration = 1.00 mol dm⁻³
• Temperature = 25°C (298 K)
• Pressure = 1 atm

When the ion concentration is changed, we no longer use E°, but rather E (non-standard electrode potential).

Effect of ion concentration on electrode potential
The position of redox equilibrium shifts when ion concentration changes, which affects the electrode potential.

This can be explained using Le-Chatelier's Principle.
• Increase in concentration of species on the left side of the half-equation → equilibrium shifts right → E becomes more positive / less negative.
• Decrease in concentration of species on the left side → equilibrium shifts left → E becomes more negative / less positive.

Zn²⁺/Zn Half-Cell
Zn²⁺₍aq₎ + 2e⁻ ⟶ Zn₍s₎

Standard E° = − 0.76 V

Table 10.1 Variation of Potential with Ion Concentration

[Fe³⁺] (mol dm⁻³) | [Fe²⁺] (mol dm⁻³) | E° (V) | Effect
> 1.00 | 1.00 | + 0.85 | More positive
< 1.00 | 1.00 | + 0.70 | Less positive
1.00 | > 1.00 | + 0.70 | Less positive
1.00 | < 1.00 | + 0.85 | More positive

Zn²⁺₍aq₎ + e⁻ ⟶ Zn₍s₎ E° = −0.76 V

Fe³⁺/Fe²⁺ Half-Cell
Fe³⁺ + e⁻ ⇌ Fe²⁺ E° = +0.77V
1.00 mol dm⁻³ 1.00 mol dm⁻³

NERST EQUATION

Q15. Write detail note on NERST equation to explain variation of electron potential.

Ans. Nernst equation
The effect of concentration and temperature on the value of E°cell can be deduced using the Nernst equation.

Consider a cell made up from a silver/silver ion electrode and a copper/copper(II) ion electrode. The reaction taking place is:
Cu₍s₎ + 2Ag⁺₍aq₎ ⟶ Cu²⁺₍aq₎ + 2Ag₍s₎ E° = + 0.46V

When E°cell is plotted against concentration a graph is obtain as shown in figure.

A graph is plotted the value of E°cell (non-standard conditions for the cell as a whole) against the logarithm of the silver ion concentration.

The above redox cell reaction makes the value of E°cell more positive.

For a given electrode Cu₍s₎ / Cu²⁺₍aq₎
The relationship is:

E = E° + (RT/zF) ln [Oxidized form]
[Reduced from]

The Nernst equation relates the electrode potential (E) of a half-cell or full electrochemical cell under non-standard conditions to:
• The standard electrode potential (E°)
• The temperature
• The concentration of ions involved in the reaction

Where,
E is the electrode potential under non-standard conditions
E° is the standard electrode potential
R is the gas constant, 8.314 J K⁻¹ mol⁻¹
T is the kelvin temperature
z is the number of electrons transferred in the reaction
F is the value of the Faraday constant in C mol⁻¹
ln is the natural logarithm
[oxidized] refers to the concentration of the oxidized form in the half-equation
[reduced] refers to the concentration of the reduced form in the half-equation.

For a metal/metal ion electrode, we can simplify this equation in three ways.
The natural logarithm, ln , is related to log to the base 10 by the relationship
ln x = 2.303 log10 x

At standard temperature the values of R, T and F are constant
The equation then becomes

E=E°+ 0.059/z log [oxidizedform]
[reducedform]

Ans. Nernst equation
The effect of concentration and temperature on the value of E°cell can be deduced using the Nernst equation.

Consider a cell made up from a silver/silver ion electrode and a copper/copper(II) ion electrode. The reaction taking place is:
Cu₍s₎ + 2Ag⁺₍aq₎ ⟶ Cu²⁺₍aq₎ + 2Ag₍s₎ E° = + 0.46V

When E°cell is plotted against concentration a graph is obtain as shown in figure.

A graph is plotted the value of E°cell (non-standard conditions for the cell as a whole) against the logarithm of the silver ion concentration.

The above redox cell reaction makes the value of E°cell more positive.

For a given electrode Cu₍s₎ / Cu²⁺₍aq₎
The relationship is:

E = E° + (RT/zF) ln [Oxidized form]
[Reduced from]

The Nernst equation relates the electrode potential (E) of a half-cell or full electrochemical cell under non-standard conditions to:
• The standard electrode potential (E°)
• The temperature
• The concentration of ions involved in the reaction

Where,
E is the electrode potential under non-standard conditions
E° is the standard electrode potential
R is the gas constant, 8.314 J K⁻¹ mol⁻¹
T is the kelvin temperature
z is the number of electrons transferred in the reaction
F is the value of the Faraday constant in C mol⁻¹
ln is the natural logarithm
[oxidized] refers to the concentration of the oxidized form in the half-equation
[reduced] refers to the concentration of the reduced form in the half-equation.

For a metal/metal ion electrode, we can simplify this equation in three ways.
The natural logarithm, ln , is related to log to the base 10 by the relationship
ln x = 2.303 log10 x

At standard temperature the values of R, T and F are constant
The equation then becomes

E=E°+ 0.059/z log [oxidizedform]
[reducedform]

Sample problem 10.6
What is the electrode potential of a Zn electrode dept in a solution containing Zn²⁺ ions (0.1M), the e° value of Zn²⁺ / ZnO is -0.76V

Using Half equation and E°,
Zn²⁺₍aq₎ + 2e⁻ ⟶ Zn₍s₎ E° = - 0.76V

E = (−0.76) + (8.31 × 298)/(2 × 96500) ln(0.1)

E = -0.76 + (-0.029) = -0.78V

Since, c0ncenration of Zn²⁺ is decreased to shift equilibrium backward and electrode potential decreases to -0.78V

Ans. Activity series
The activity series of metals is a ranking of metals based on their reactivity, particularly their tendency to lose electrons and undergo oxidation.

(i) Highly reactive metals
• Metals higher in the activity series have more negative standard reduction potentials.
• They are more likely to lose electrons and undergo oxidation.
• Such metals act as reducing agents.
• Metals at the top of the series are the most reactive with water, oxygen and acid.

Example Alkali metals like cesium (Cs) and sodium (Na) are highly reactive and oxidize readily, reacting vigorously with substances like water of to form ions and release hydrogen gas.

Reaction of sodium with water
2Na + 2H₂O ⟶ 2NaOH + H₂ ↑

(ii) Least reactive metals
• Metals lower in the activity series, such as noble metals like gold and silver, do not oxidize easily.
• Metals at the bottom of the series are the least reactive with water, oxygen and acid.

Example Metals such as gold, platinum and silver are least active metals because they are lowest in the series from metals.

Reactivity series
• The position of a metal in the activity series directly correlates with its ease of oxidation and overall reactivity.
• Activity series of metals is actually reactivity series.
• The activity series can be understood in terms of standard reduction potentials.
• Metals with more negative potentials are more reactive and oxidize more readily.
• Metals with more positive potentials are less reactive and oxidize less readily as given in Table 10.2.

Conclusion
The activity series helps predict:
• Metal displacement reactions
• Reactivity with water and acids
• Whether a redox reaction will occur

It's a practical tool in electrochemistry, metallurgy, and industrial extraction of metals.

Ans. Feasibility of redox reactions from activity series data
The activity series ranks metals by their reactivity, from most reactive (top) to least reactive (bottom).

A metal higher in the series can displace a metal lower in the series from its salt solution.
This allows us to predict if a redox reaction is feasible.

Rule for feasibility • A more reactive metal (higher in the series) can reduce the ion of a less reactive metal.
• If the overall cell potential ( E°cell ) is positive, the reaction is feasible.
• If E°cell is negative, the reaction is not feasible.

Step-by-step process to check feasibility:
(i) Write half-reactions for oxidation and reduction.
(ii) Look up standard electrode potentials (E° values).
(iii) Calculate E°cell:
Interpret the result:
• E°cell > 0 then reaction is feasible
• E°cell < 0 then reaction is not feasible

Consider an example of Fe²⁺ and Ag⁺ from activity series.

Half-Reactions Fe²⁺ + 2e⁻ ⇌ Fe₍s₎ E°(ox) = -0.44V

Ag⁺ + 1e⁻ ⇌ Ag₍s₎ E°(ox) = +0.80V

Overall reaction Fe₍s₎ + 2Ag⁺₍aq₎ ⟶ Fe²⁺₍aq₎ + 2Ag₍s₎

Cell potential
E°cell = 0.80 − (−0.44) = +1.24V

The cell potential is positive so the reaction is feasible.

Consider another example from the activity series of metals Al³⁺ and Ag⁺.

Al³⁺ + 3e⁻ ⇌ Al₍s₎ E°(red) = −1.66V

Ag⁺ + e⁻ ⇌ Ag₍s₎ E°(ox) = +0.80V

The overall reaction is
Al³⁺₍aq₎ + Ag₍s₎ ⟶ Al₍s₎ + Ag⁺₍aq₎

Cell potential E°cell = −1.66 +0.80 = −0.86V

The cell potential is negative so the reaction is not feasible.

Conclusion • Use the activity series or E° values to predict feasibility.
• If a metal is higher in the series, it can replace a metal lower in the series.
• Positive E°cell confirms the reaction can proceed spontaneously.

Ans. The word 'photovoltaic' is composed of photo (light) and volt (electrical potential).
The cell which converts light energy into electrical energy is called photovoltaic cell.

Principle of photovoltaic cells
Definition: A photovoltaic (PV) cell is a device that converts light energy directly into electrical energy using the photovoltaic effect.

Photovoltaic Effect
• The photovoltaic effect is the basic principle behind the working of solar cells.
• It is the generation of voltage and current in a material when exposed to light.

Working

1. Light Absorption
• When sunlight (photons) strikes the surface of a semiconductor, it transfers energy to electrons.

2. Excitation of Electrons
• The energy from photons excites electrons, allowing them to jump from the valence band to the conduction band.
• This creates free electrons (negative charge) and holes (positive charge).

3. PN Junction Role
• A PN junction (junction between p-type and n-type semiconductors) is built inside the PV cell.
• The junction helps in separating electrons and holes, preventing recombination.

4. Electric Field Generation
• The electric field at the PN junction drives electrons toward the n-side and holes toward the p-side.
• This movement creates a potential difference (voltage).

5. Current Flow
• When the external circuit is connected, electrons flow through the circuit, producing electric current.

Merits of photovoltaic cell as sustainable source of energy
• The photovoltaic effect continues as long as light is present, making it a sustainable method for generating electricity.
• A photovoltaic system is a renewable energy source that converts sunlight into electrical energy.
• The photovoltaic systems use the sun's energy, making them a sustainable energy source that is independent of fossil fuels.

Ans. Biochemical Oxygen Demand (BOD)
• Definition: BOD is the amount of oxygen consumed by microorganisms to decompose organic matter in water over a period of 5 days at 25°C.
• Significance: It reflects the level of organic pollution in water—higher BOD means more pollution.

Measurement Calculated by measuring the difference in dissolved oxygen (DO) in a water sample at the start and after 5 days.

Winkler Method for DO Measurement
Principle: DO in the water oxidizes potassium iodide (KI) to release iodine.

Process (i) Add MnSO₄ and alkaline KI to the sample.
(ii) Oxygen reacts, forming a brown precipitate of MnO(OH)₂.
(iii) Acidify the sample → iodine is released.
(iv) Titrate iodine with sodium thiosulfate using starch as an indicator (blue color indicates presence of iodine). The blue color disappears—this marks the amount of DO. More the amount of iodine formed, more the value of DO.