FSc Part 1 Chemistry Chapter 10 Complete Guide Book — Electrochemistry

Electrochemistry is the branch of chemistry which deals with the interconversion of electrical and chemical energy.

Oxidation is a process involving loss of electron or electrons: Fe²⁺ → Fe³⁺ + e⁻; Zn⁰ → Zn²⁺ + e⁻. Here Fe²⁺ and Zn⁰ have lost electrons, therefore they are oxidized.

Reduction is a process involving gain of electron or electrons: Cl⁰ + e⁻ → Cl⁻¹; Cu²⁺ + 2e⁻ → Cu⁰. Here Cl⁰ and Cu²⁺ have gained electrons, therefore they are reduced.

Oxidation and reduction always take place together. The reactions in which this happens are called redox reactions. There are two ways of finding out whether a substance has been oxidized or reduced during a chemical reaction: electron transfer, and changes in oxidation number.

📌 Interesting Information (book's side note): Photosynthesis is a redox reaction which provides food for the entire planet, and another one is respiration that keeps us alive, both are redox reactions.

The oxidation number is also called oxidation state. An oxidation number is a number given to each atom or ion in a compound that shows its degree of oxidation. It is the apparent charge on an atom (per atom) of an element in a molecule or an ion. Oxidation number can be positive, negative or zero. The + or − sign must always be included. Higher positive oxidation number means that an atom or ion is more oxidized; higher negative oxidation number means that an atom or ion is more reduced.

10.2.1 Oxidation number rules: (i) The oxidation number of any uncombined element is zero, e.g. each atom in S₈, H₂, Cl₂, Na, Mg and Zn is zero. (ii) In either a compound or an ion, the more electronegative element is given the negative oxidation number. In HF, fluorine has −1. (iii) In compounds, many atoms or ions have fixed oxidation numbers: group 1 alkali metals are always +1, group 2 alkaline earth elements are always +2, group 3 elements +3, group 17 halogens in binary compounds are always −1, hydrogen is +1 (except in metal hydrides, such as NaH, where it is −1), oxygen is −2 (except in peroxides, where it is −1, and in F₂O, where it is +2). (iv) The oxidation number of an element in a mono-atomic ion is always the same as the charge on the ion, e.g. Cl⁻ is −1, Al³⁺ is +3. (v) In compounds, the algebraic sum of the oxidation numbers of all the elements is zero, e.g. HCl and NaCl. (vi) In polyatomic ions, the algebraic sum of oxidation number equals the charge on the ion, e.g. NH₄⁺, CO₃²⁻ and SO₄²⁻. (vii) For transition metals and many non-metals, oxidation numbers cannot be fixed as they show variable oxidation states; therefore their oxidation numbers are always determined by calculations.

10.2.2 Finding oxidation number in a compound or a radical: A metal always has the positive oxidation number and non-metals have the negative oxidation state. For example, in sodium sulfide, Na₂S: 2(+1) + S = 0, so S = −2. In compounds of two different non-metals, the sign depends on electronegativity; the most electronegative element is given the negative sign.

Sample Problem 10.1: Oxidation number of S in sulfur dioxide, SO₂: S + 2(−2) = 0, so S = +4. Sample Problem 10.2: Oxidation number of sulfur in SO₄²⁻: [Ox. No. of S] + 4(−2) = −2, so S + (−8) = −2, S = +6.

📌 Keep in Mind! (book's side note): Hydrogen shows −1 oxidation state with all metals and +1 with all non-metals. Oxygen shows negative oxidation state with all metals and with all non-metals except fluorine. Oxygen shows −2, −1 and −1/2 in H₂O, Na₂O₂ and KO₂ respectively.

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 of the same element. This type of reaction involves the simultaneous oxidation and reduction of the same element.

For example, the decomposition of hydrogen peroxide (H₂O₂) to form water (H₂O) and oxygen (O₂) is a disproportionation reaction: 2H₂O_2(l) → 2H₂O_(l) + O_2(g). Oxidation number of oxygen in H₂O₂ = −1; in H₂O = −2 (oxygen is reduced from −1 to −2); in O₂ = 0 (oxygen is oxidized from −1 to 0).

Similarly: 6NaOH_(aq) + 3Cl_2(g) → 5NaCl_(aq) + NaClO_3(aq) + 3H₂O. Oxidation number of chlorine atoms in Cl₂ = 0; in NaCl = −1 (Cl is reduced from 0 to −1); in NaClO₃ = +5 (Cl is oxidized from 0 to +5).

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, while a reducing agent (atom, ion or molecule) is that substance which reduces some other substance, and is itself oxidized to a higher oxidation state by losing one or more electrons.

If an element X gains one or more electrons and is converted into X⁻ anion, the element X is said to be acting as an oxidizing agent, since it withdraws an electron from the other atom; in this process oxidation number of X has decreased: X + e⁻ → X⁻, [Ox. No. = 0] → [Ox. No. = −1], higher oxidation state → lower oxidation state. For example, Cl + e⁻ → Cl⁻.

On the other hand, if an element M loses one electron and is converted to M⁺ cation, the element M is said to be acting as a reducing agent, since it is providing electrons; it gets higher oxidation state: M → M⁺ + e⁻, [Ox. No. = 0] → [Ox. No. = +1], lower oxidation number → higher oxidation number. For example, Na → Na⁺ + e⁻.

Memory rule: the oxidizing agent is itself reduced (its oxidation number falls); the reducing agent is itself oxidized (its oxidation number rises). Do not mix the two up — the agent does the opposite of its own change to the other substance.

Carry out the following steps for balancing of redox equations by oxidation number method: (i) Write down the skeleton equation of the redox reaction under consideration. (ii) Identify the elements which undergo a change in their oxidation number during the reaction. (iii) Record the oxidation number above the symbols of the elements which have undergone a change in the oxidation number. (iv) 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 increase or decrease in the oxidation number, i.e. electrons gained or lost by multiplying with a suitable digit. (vi) Balance the rest of the equation by inspection (hit and trial) method.

📌 Keep in Mind! (book's side note): (a) If an element in a species undergoes only increase in oxidation number or decrease in oxidation number in a reaction, the specie containing such element is written once on the RHS of the equation, e.g. Zn_(s) + HCl_(aq) → ZnCl_2(aq) + H_2(g). (b) If an element in a species undergoes both increase in oxidation number or decrease in oxidation number simultaneously in a reaction (as in case of self-redox or disproportionation reaction), the specie containing such element is written twice on the RHS of the equation. First for increase in oxidation number while second is written for decrease in oxidation number, e.g. Cl₂ + NaOH + Cl₂ → NaCl + H₂O + NaOCl. (c) If an element in a species undergoes increase or decrease in oxidation number as well as no change in oxidation number in a reaction, the specie containing such element is also written twice. First, for change in oxidation number while second is for no change in oxidation number, e.g. HNO₃ + Cu + HNO₃ → Cu(NO₃)₂ + H₂O + NO₂; HCl + K₂Cr₂O₇ + HCl → Cl₂ + KCl + CrCl₃ + H₂O.

Sample Problem 10.4 (as numbered in the book): Balance K₂Cr₂O₇ + HCl → KCl + CrCl₃ + Cl₂ + H₂O by oxidation number method. Cr undergoes a change from +6 to +3 (reduced, acts as oxidizing agent) while Cl undergoes a change from −1 to 0 (oxidized, acts as reducing agent). Total decrease = 2 Cr × 3 = 6e⁻; total increase = 2e⁻ per Cl₂, so multiply Cl₂ by 3 and Cr by 2: K₂Cr₂O₇ + HCl → 3Cl₂ + 2KCl + 2CrCl₃ + H₂O. Balance the rest by inspection: K by multiplying KCl by 2 (already counted), Cl atoms by multiplying HCl by 14, oxygen by multiplying H₂O by 7: K₂Cr₂O₇ + 14HCl → 3Cl₂ + 2KCl + 2CrCl₃ + 7H₂O.

📌 Did you know? (book's side note): Sequence of the balancing the chemical equation using inspection (hit and trial) method. Firstly, balance atoms of all elements except nitrogen, oxygen and hydrogen. Secondly, balance atoms of nitrogen (if in the equation). Thirdly, balance atoms of oxygen (if in the equation). Lastly, balance atoms of hydrogen (if in the equation).

10.6 Electrolytic cell: An electrolytic cell is a device that converts electrical energy into chemical energy through a process called electrolysis. It consists of two electrodes, typically made of metal or another conductive material, which are immersed in an electrolyte. When a direct electric current is applied, one electrode becomes negatively charged (the cathode) and the other positively charged (the anode).

In an electrolytic cell, positive ions in the electrolyte migrate towards the cathode, where they gain electrons (a reduction process), while negative ions move towards the anode, where they lose electrons (an oxidation process). This movement of ions and transfer of electrons results in chemical changes at the electrodes, effectively converting electrical energy into chemical energy.

Electrolytic cells are used in the electrolysis of sodium chloride to produce sodium metal and chlorine gas, the refining and plating of metals, and the production of chemicals like caustic soda.

Points to remember: in the electrolysis cell, the electrolyte is the compound that is ionized; it is either a molten ionic compound or a concentrated aqueous solution of ions. The electrodes are made from either carbon (graphite) or metal, which conduct electricity to and from the electrolyte; the anode is the positive electrode, the cathode is the negative electrode. The power supply must be direct current. The actual structure of the cell will vary according to the element extraction.

10.7 Redox reactions in electrolysis: During electrolysis, the positive ions (cations) move to the cathode. When they reach the cathode, they gain electrons from the cathode, e.g. Cu²⁺_(aq) + 2e⁻ → Cu_(s); 2H⁺_(aq) + 2e⁻ → H_2(g). Gain of electrons is reduction and it always occurs at the cathode. If metal atoms are formed, they may be deposited as a layer of metal on the cathode. Alternatively, they may form a molten layer in the cell. If hydrogen gas is formed, it bubbles off.

The negative ions (anions) move to the anode. When they reach the anode, they lose electrons to the anode. Loss of electrons is oxidation. Oxidation always occurs at the anode, e.g. 2Cl⁻_(aq) → Cl_2(g) + 2e⁻; 4OH⁻_(aq) → O_2(g) + 2H₂O_(l) + 4e⁻.

Electrolysis is a redox reaction. For example, when molten zinc chloride (ZnCl₂) is electrolyzed, the electrode reactions are: Cathode: Zn²⁺_(l) + 2e⁻ → Zn_(s) (Reduction); Anode: 2Cl⁻_(l) → Cl_2(g) + 2e⁻ (Oxidation). The electron loss at the anode balances the electron gain at the cathode. The overall reaction is ZnCl_2(l) → Zn_(s) + Cl_2(g).

10.8 Mass of a substance deposited during electrolysis: The mass of a substance produced at an electrode during electrolysis is proportional to the time over which a constant electric current passes and the strength of the electric current. Combining current and time, we get the relationship: Q = I × t, where Q = charge (in coulombs, C), I = current (in amperes, A), t = time (in seconds, s).

The mass of a substance produced at (or removed from) an electrode during electrolysis is proportional to the quantity of electricity (in coulombs) which passes through the electrolyte. The quantity of electricity is often expressed in terms of a unit called the Faraday (symbol F). 1 Faraday is the quantity of electric charge carried by 1 mole of electrons or 1 mole of singly charged ions. Its value is 96500 C mol⁻¹.

During the electrolysis of silver nitrate solution, silver is deposited at the cathode: Ag⁺_(aq) + e⁻ → Ag_(s) (1 mol, 1 mol, 1 mol). 1 Faraday of electricity (96500 C) is required to deposit 1 mole of silver (107.9 g). During the electrolysis of copper(II) sulfate solution, copper is deposited at the cathode: Cu²⁺_(aq) + 2e⁻ → Cu_(s) (1 mol, 2 mol, 1 mol). The equation shows that 2 moles of electrons are needed to produce 1 mole of copper from Cu²⁺ ions. So, it requires 2 Faradays of electricity (2 × 96500 C) to deposit 1 mole of copper (63.5 g).

Number of Faraday = number of moles of electrons gained or lost.

10.9 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.

Sample Problems 10.5: Calculate the mass of lead (Pb) deposited at the cathode during electrolysis when a current of 1.50 A flows through molten lead(II) bromide (PbBr₂) for 20.0 min (relative atomic mass, A_r value: [Pb] = 207; F = 96500 C mol⁻¹). Solution: Step 1 — half-equation: Pb²⁺_(aq) + 2e⁻ → Pb_(s). Step 2 — coulombs to deposit 1 mole: 2 moles of electrons per mole of Pb formed = 2F = 2 × 96500 = 193000 C mol⁻¹. Step 3 — charge transferred: Q = I × t = 1.50 × 20 × 60 = 1800 C. Step 4 — mass by simple proportion using the relative atomic mass: 193000 C deposits 1 mole Pb, which is 207 g Pb, so 1800 C deposits = 1.93 g Pb (207/193000 × 1800).

The Avogadro constant, N_A, is the number of specified particles (atoms, molecules, or ions) in 1 mole. An electrolytic method can be used to find a value for the Avogadro constant by calculating the charge associated with 1 mole of electrons: N_A = (Charge on 1 mole of electrons) ÷ (Charge on 1 electron).

The charge on the electron can be calculated by experiment. The results show us that the charge on the electron is approximately 1.60 × 10⁻¹⁹ C.

10.10.1 Finding the charge on 1 mole of electrons: The charge on 1 mole of electrons can be found from a simple electrolytic experiment. The apparatus is a beaker of aqueous copper(II) sulfate with a pure copper anode and a pure copper cathode, connected through an ammeter, a battery and a variable resistor. The procedure is to weigh the pure copper anode and pure copper cathode separately, arrange the apparatus during the electrolysis of aqueous copper(II) sulfate, use the variable resistor to keep the passing current constant about 0.2 amperes, pass a constant electric current for a measured time interval, e.g. 40 min, remove the cathode and anode and wash and dry them with distilled water and then with propanone, and reweigh the cathode and anode. The cathode increases in mass because copper is deposited. The anode decreases in mass because the copper goes into solution as copper ions. The decrease in the mass of the anode is measured. This is preferred because the copper does not always stick to the cathode very well.

10.10.2 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 0.136 g, so 0.136 g of copper requires 408 C; 1 g of copper requires 408/0.136 C; 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 = 95,250 C. If the charge on one electron is 1.60 × 10⁻¹⁹ C, then (book's working) 95250 ÷ (1.60 × 10⁻¹⁹) gives 5.95 × 10²³ — the book prints this step in a compressed form and states the result is in good agreement with the accurate value of 6.02 × 10²³ mol⁻¹.

Note the logic honestly: the Faraday constant from this experiment is about 95,250 C mol⁻¹ (close to the accepted 96,500 C mol⁻¹), and N_A = F ÷ e = 96500 ÷ (1.60 × 10⁻¹⁹) ≈ 6.03 × 10²³ mol⁻¹. The book's own numbers (190500 C for 63.5 g Cu) are experimental, so the small difference from 96,500 is experimental error, not a new value of F.

During the redox processes, a redox equilibrium exists between two chemically related species that are in different oxidation states. When a metal is put into a solution of its ions, an electric potential (voltage) is established between the metal and the metal ions in solution. This is called electrode potential and it indicates the ease of oxidation or reduction of a substance. For example, when a copper rod is placed in an aqueous solution of its ions, the following equilibrium exists: Cu²⁺_(aq) + 2e⁻ ⇌ Cu_(s).

There are two opposing reactions in this equilibrium: (a) Metal atoms from the electrode entering the solution as metal ions. This leaves electrons on the electrode: Cu_(s) → Cu²⁺_(aq) + 2e⁻. (b) Ions in solution accepting electrons from the metal electrode and getting deposited as metal atoms on the surface of the electrode: Cu²⁺_(aq) + 2e⁻ → Cu_(s). The redox equilibrium is established when the rate of electron gain equals the rate of electron loss.

For unreactive metals such as copper, if this equilibrium is compared with the equilibrium set up by other metals, the equilibrium Cu²⁺_(aq) + 2e⁻ ⇌ Cu_(s) lies further to the right in the sense used by the book: Cu²⁺ ions are therefore relatively easy to reduce. They gain electrons readily to form copper metal. For reactive metals such as vanadium, the equilibrium lies further over to the left: V²⁺_(aq) + 2e⁻ ⇌ V_(s). V²⁺ ions are therefore relatively difficult to reduce. They gain electrons much less readily by comparison. The position of equilibrium differs for metals placed in solutions of their ions.

10.11.1 Electrical double layer: It is thought that the absolute electrode potentials cannot be measured due to the formation of an electrical double layer, when an element is placed in a solution of its ions. For example, when zinc is placed in a solution containing zinc ions, a few number of zinc atoms on the surface of the metal is converted to zinc ions, which go into solution. This leaves an excess of electrons on the surface of the zinc electrode. The solution around the metal now has excess Zn²⁺ ions. Some of these cations near the surface of the zinc are attracted to the surface. So, an electrical double layer is formed. This buildup of charge causes an electric potential (voltage) between the metal and the metal ions in solution. Due to the formation of an electrical double layer, the potential cannot directly be measured. But the difference in potential between the metal/metal ion system and another system can be measured. We call this value the electrode potential, E. Electrode potential is measured in volts. The system we use for comparison is the standard hydrogen electrode (SHE).

10.12 Standard hydrogen electrode (SHE): The standard hydrogen electrode is one of several types of half-cell that can be used as reference electrode. This electrode consists of: hydrogen gas at 101 kPa pressure, in equilibrium with H⁺ ions of concentration 1.00 mol dm⁻³; a platinum electrode covered with platinum black in contact with the hydrogen gas and the H⁺ ions.

The platinum black is finely divided platinum, which allows close contact of hydrogen gas and H⁺ ions in solution so that equilibrium between H₂ gas and H⁺ ions is established quickly. Platinum electrode is inert. Standard electrode potential E° values for all half-cells are measured relative to this electrode. When connected to another half-cell, the value read on the voltmeter gives the standard electrode potential for that half-cell. The half-equation for the hydrogen electrode can be written: 2H⁺_(aq) + 2e⁻ ⇌ H_2(g), E° = 0.00 V. It means that hydrogen gas and hydrogen ions have equal tendency to lose and gain electrons.

10.13 Standard electrode potential (E°): The voltage of an electrochemical cell will also depend on concentration, temperature, and pressure of the gas. Therefore, we should use standard conditions when comparing electrode potentials. These are: concentration of ions at 1.00 mol dm⁻³; a temperature of 25 °C (298 K); any gases should be at a pressure of 1 atmosphere (101 kPa); the value of the electrode potential of the half-cell is measured relative to the standard hydrogen electrode. Under these conditions, the electrode potential we measure is called the standard electrode potential. This has the symbol, E°. It is spoken of as 'E standard'. "The standard electrode potential for a half-cell is the voltage measured under standard conditions with a standard hydrogen electrode (SHE) as the other half-cell."

10.14 Measuring standard electrode potentials: There are three main types of half-cell whose E° value can be obtained when connected to a standard hydrogen electrode (SHE): metal/metal ion half-cell (Cu²⁺/Cu⁰); non-metal/non-metal ion half-cell (Cl¹⁻/Cl₂) (Pt or C electrode is used); ion/ion half-cell (Fe²⁺/Fe³⁺) (Pt or C electrode is used).

The Cu²⁺/Cu half-cell is connected to the SHE and the voltage measured. The voltage is +0.34 V. The two half-equations are: Cu²⁺_(aq) + 2e⁻ ⇌ Cu_(s), E° = +0.34 V; H⁺_(aq) + e⁻ ⇌ 1/2 H_2(g), E° = 0.00 V. The E° values show us that Cu²⁺ ions are easier to reduce than H⁺ ions (they have a more positive E° value). Cu²⁺ ions are more likely to gain electrons than H⁺ ions. So Cu²⁺ ions will accept electrons from the H⁺/H₂ half-cell and H₂ will lose electrons to the Cu²⁺/Cu half-cell.

The Zn²⁺/Zn half-cell voltage is −0.76 V. The two half-equations are: 2H⁺_(aq) + 2e⁻ ⇌ H_2(g), E° = 0.00 V; Zn²⁺_(aq) + 2e⁻ ⇌ Zn_(s), E° = −0.76 V. The E° values show us that Zn²⁺ ions are more difficult to reduce than H⁺ ions (they have a more negative E° value). Zn²⁺ ions are less likely to gain electrons than H⁺ ions. So, Zn will lose electrons to the H⁺/H₂ half-cell and H⁺ ions will gain electrons from the Zn²⁺/Zn half-cell. From these two examples, we can see that: reduction takes place at the positive terminal of the cell, e.g. in the Zn²⁺/Zn; H⁺/H₂ cell: 2H⁺_(aq) + 2e⁻ → H_2(g); oxidation takes place at the negative terminal of the cell, e.g. Zn_(s) → Zn²⁺_(aq) + 2e⁻.

10.15 Electrochemical cell (galvanic 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. A galvanic cell is designed to take advantage of spontaneous transfer of electrons. For example: Cu²⁺/Cu // Zn²⁺/Zn, E° = 1.10 V. If we connect two half-cells together, we have made an electrochemical cell. We can measure the voltage of this cell. A Cu²⁺/Cu half-cell connected to a Zn²⁺/Zn half-cell makes a complete electrochemical cell.

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. 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.

The voltages for the half-cells can be represented by the following half-equations: Cu²⁺_(aq) + 2e⁻ ⇌ Cu_(s), E° = +0.34 V; Zn²⁺_(aq) + 2e⁻ ⇌ Zn_(s), E° = −0.76 V. The relative values of these voltages tell us that Zn²⁺ ions are more difficult to reduce than Cu²⁺ ions. At the anode, zinc undergoes oxidation, losing electrons to form zinc ions: Zn_(s) → Zn²⁺_(aq) + 2e⁻. These electrons travel through an external circuit to the cathode, where copper ions gain electrons and are reduced to form copper metal: Cu²⁺_(aq) + 2e⁻ → Cu_(s). The overall cell reaction can be represented as: Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s). This 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: E°_cell = E°_red + E°_oxi = +0.34 − (−0.76) = +1.10 V (book's form: reduction potential plus oxidation potential; equivalently E°_cell = E°_cathode − E°_anode). In a Cu-Zn galvanic cell, zinc (Zn) serves as the anode, and copper ions (Cu²⁺) in solution serve as the cathode. 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.

10.16 Applications of E° values — 10.16.1 Direction of electron flow: We can deduce the direction of electron flow in the wires in the external circuit by comparing the E° values for the two half-cells which make up the electrochemical cell. For example, Ag⁺_(aq) + e⁻ ⇌ Ag_(s), E° = +0.80 V; Zn²⁺_(aq) + 2e⁻ ⇌ Zn_(s), E° = −0.76 V. The relative values tell us that Zn²⁺ ions are more difficult to reduce than Ag⁺ ions. So, Zn metal will lose electrons to the Ag⁺/Ag half-cell and Ag⁺ ions will accept electrons from the Zn²⁺/Zn half-cell. The electrons move through the wires in the external circuit. They do not travel through the electrolyte solution. So, the electron flow is from the Zn²⁺/Zn half-cell to the Ag⁺/Ag half-cell. In other words, the flow is from the negative pole to the positive pole. Remember that the more positive pole attracts the negative electrons.

10.16.2 Feasibility of a reaction using E° values: Standard electrode potential values, E°, give us a measure of how easy or difficult it is to oxidize or reduce a species. The E° values are listed in order of increasingly negative values. For each half-equation, the more oxidized form is on the left and the more reduced form is on the right. The more positive the value of E°, the greater the tendency for the half-equation to proceed in the forward direction, and the easier it is to reduce the species on the left of the half-equation. The less positive the value of E°, the greater the tendency for the half-equation to proceed in the reverse direction, and the easier it is to oxidize the species on the right of the half-equation.

To check the feasibility of reaction between Mg²⁺ and Ag: Mg²⁺ + Ag → ? We need half-cell equations and E° values: Mg²⁺ + 2e⁻ ⇌ Mg, E° = −2.32 V; Ag⁺ + 1e⁻ ⇌ Ag, E° = +0.80 V. Mg²⁺ are supposed to be reduced here and Ag is supposed to get oxidized, so calculating E°_cell gives: E°_cell = E°_red − E°_ox = (−2.32) − (+0.80) = −3.12 V by the strict cathode-minus-anode rule (the book prints −1.52 V for this step, which does not follow from its own numbers — see the note in the exercise/report; the conclusion is unchanged). 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: E°_cell = E°_red − E°_ox = (+0.80) − (−2.32) = +3.12 V by the same rule (the book prints +1.52 V, again half the strict value — conclusion unchanged). Therefore, a feasible reaction is 2Ag⁺ + Mg_(s) → Mg²⁺_(aq) + 2Ag_(s) (book's equation; its printed +1.52 V is corrected here to +3.12 V from the book's own E° values).

10.16.3 Oxidising and reducing agents using E° values: In the reaction between Cu and Zn²⁺, Cu will not reduce Zn²⁺ ions to Zn. So how can we reduce Zn²⁺ ions? The answer is to react the Zn²⁺ ions with a stronger reducing agent, which should have an E° value more negative than the E° value for Zn²⁺/Zn. We see that the half-equation Mg²⁺/Mg has a more negative E° value. So, Mg is a suitable reducing agent: Zn²⁺_(aq) + e⁻ → Zn_(s), E° = −0.76 V; Mg²⁺_(aq) + 2e⁻ → Mg_(s), E° = −2.38 V (book prints −2.38 V here; its table on p. 229 gives Mg −2.37 V — a 0.01 V rounding difference). Zn²⁺ is the better oxidizing agent. It is more likely to accept electrons than Mg²⁺ ions. Mg is the better reducing agent. It is more likely to release electrons than Zn. Compounds can also act as oxidizing and reducing agents. For example KMnO₄ (+0.54 V as printed in the book) and K₂Cr₂O₇ (+1.33 V) are oxidizing agents because of their higher E° values, whereas KI and FeSO₄ are reducing agents due to their lower E° values.

10.17 Variation of E° 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. Under the non-standard conditions we use the symbol E for the electrode potential. Let us take an example of a metal/metal ion equilibrium, and the book's Table 10.1 for the Fe³⁺/Fe²⁺ system (standard E° = +0.77 V at 1.00 mol dm⁻³ each): [Fe³⁺] > 1.00, [Fe²⁺] = 1.00 → E = +0.85 V, more positive; [Fe³⁺] < 1.00, [Fe²⁺] = 1.00 → E = +0.70 V, less positive; [Fe³⁺] = 1.00, [Fe²⁺] > 1.00 → E = +0.70 V, less positive; [Fe³⁺] = 1.00, [Fe²⁺] < 1.00 → E = +0.85 V, more positive.

Le-Chatelier's principle can be applied to redox equilibria. If we increase the concentration of the species on the left of the equation, the position of equilibrium will shift to the right. So, the value of E becomes more positive / less negative with change in concentration of ions. If two different ions are present in the half-cell, we have to consider both ions. Let us take the equilibrium between Fe³⁺ ions and Fe²⁺ ions as an example: Fe³⁺ + e⁻ ⇌ Fe²⁺, E° = +0.77 V, both at 1.00 mol dm⁻³.

10.18 Nernst equation: If we 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° = +0.46 V. When E°_cell is plotted against concentration we obtain a graph: E_cell rises with log [Ag⁺] towards 0.46 V at standard concentration. Note that we have 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 as silver ion concentration increases.

The effect of concentration and temperature on the value of E_cell can be deduced using the Nernst equation. For a given electrode, e.g. a Cu²⁺/Cu⁰_(aq) electrode, the relationship is: E = E° + (RT/zF) ln ([oxidized form]/[reduced form]), 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.

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, so the equation then becomes: E = E° + (0.059/z) log ([oxidized form]/[reduced form]).

Sample problem 10.5 (second problem so numbered in the book — duplicate numbering, see note): What is the electrode potential of a Zn electrode dipped in a solution containing Zn²⁺ ions (0.1 M)? The E° value of Zn²⁺/Zn is −0.76 V. Using half-equation and E°: Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V. E = (−0.76) + (8.31 × 298)/(2 × 96500) ln (0.1) = −0.76 + (−0.029) = −0.78 V as printed in the book. (Exact arithmetic gives −0.76 − 0.0295 = −0.7895 ≈ −0.79 V; the book's −0.78 V is its rounding of the 0.029 term.) Since concentration of Zn²⁺ is decreased, the equilibrium shifts backward and electrode potential decreases to about −0.78/−0.79 V.

10.19 Activity series of metals: The activity series of metals is a ranking of metals based on their reactivity, particularly their tendency to lose electrons and undergo oxidation. Metals higher in the activity series have more negative standard reduction potentials, meaning they are more likely to lose electrons and undergo oxidation. Hence such metals act as reducing agents. Metals at the top of the series are the most reactive with water, oxygen and acid. For example, alkali metals like cesium and sodium are highly reactive and oxidize readily, reacting vigorously with substances like water to form ions and release hydrogen gas.

In contrast, 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. Thus, 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.

Thus, the activity series can be understood in terms of standard reduction potentials: metals with more negative potentials are more reactive and oxidize more readily, while those with more positive potentials are less reactive and oxidize less readily as given in Table 10.2 (book, p. 229): Li −3.05, K −2.93, Ba −2.90, Sr −2.89, Ca −2.87, Na −2.71, Mg −2.37, Al −1.66, Zn −0.76, Cr −0.74, Fe −0.44, Cd −0.40, Co −0.28, Ni −0.25, Sn −0.14, Pb −0.13, H₂ 0.00 (reference), Cu +0.34, Hg +0.92, Ag +0.80, Pt +1.18, Au +1.50 (all E° in V; note the book's table order puts Hg +0.92 above Ag +0.80, out of strict numerical order). Top group (Li–Ca) reacts with cold water replacing hydrogen; Mg–Fe group reacts with steam but not cold water replacing hydrogen; Ni–Pb group does not react with water but reacts with acids replacing hydrogen; Cu–Au group is unreactive with water or acids. Trend down the series: decreasing ease of oxidation, reactivity and acting as reducing agent.

10.20 Feasibility of redox reactions from activity series or reaction data: The activity series is actually a reactivity series of metals, in order of reactivity from highest to lowest. It is used to determine the products of single displacement reactions, whereby metal M₁ will replace another metal M₂ in a solution if M₁ is higher in the series. From the activity series of metals, we can determine whether the reaction is feasible or not. We sum up the potentials of half-cell reactions; if potential of the cell is positive, the reaction is possible, if negative, the reaction is not feasible. Let us see the reactions:

Fe²⁺_(aq) + 2e⁻ ⇌ Fe_(s), E°_(ox) = −0.44 V (book's label); Ag⁺_(aq) + 1e⁻ ⇌ Ag_(s), E°_(ox) = +0.080 V as printed in the book (table value +0.80 V — a decimal slip in the printed label; note that 0.080 − (−0.44) = +0.52 V, not the book's +1.24 V, so the book's final +1.24 V only follows from the table value +0.80 V). The overall reaction is Fe_(s) + 2Ag⁺_(aq) → Fe²⁺_(aq) + 2Ag_(s), E°_(cell) = 0.80 − (−0.44) = +1.24 V with the table value. The cell potential is positive; hence the reaction is feasible.

Consider another example: from the activity series of metals, we have Al³⁺ + 3e⁻ ⇌ Al, E°_(red) = −1.66 V; Ag⁺ + 1e⁻ ⇌ Ag, E°_(red) = +0.080 V (book's printed value). Al³⁺_(aq) + Ag_(s) → Al_(s) + Ag⁺_(aq), E°_(red) = −1.66 + 0.080 = −0.86 V as printed in the book. The sum of E° values of the two half-cell reactions is negative. Hence the reaction is not feasible. (Note: the book's −0.86 V here adds the two reduction potentials instead of subtracting cathode-minus-anode; with the table value Ag +0.80 V the strict E°_cell = −1.66 − (+0.80) = −2.46 V — still negative, so the book's conclusion "not feasible" is correct even though its printed arithmetic is not the standard form.)

10.21 Photovoltaic cells: 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.

10.21.1 Principle of photovoltaic cells: The central principle behind the operation of PV cells is the photovoltaic effect, which involves the generation of voltage and electric current in a material upon exposure to light. The photovoltaic effect involves the excitation of electrons from the valence band to the conduction band in semiconductors, facilitated by the energy from photons. This process is improved by the presence of a PN junction, which separates the electrons and holes, creating a voltage that drives an electric current.

10.22.2 (as numbered in the book) Merits of photovoltaic cell as sustainable source of energy: (i) The photovoltaic effect continues as long as light is present, making it a sustainable method for generating electricity. (ii) A photovoltaic system is a renewable energy source that converts sunlight into energy. (iii) The photovoltaic systems use the sun's energy, making them a sustainable energy source that is independent of fossil fuels.

10.22 Winkler method, BOD and DO: Biochemical oxygen demand (BOD) is the capacity of organic matter in natural water to consume oxygen within a period of five days. The value of BOD is the amount of oxygen consumed as a result of biological oxidation of dissolved organic matter in the sample. The oxidation reaction is catalyzed by microorganisms which are already present in the natural water. It is measured experimentally through Winkler method by calculating the concentration of oxygen in the beginning and at the end of five day period, in which a sealed water sample is maintained in the dark at constant temperature at 25 °C.

Dissolved oxygen (DO) is defined as the amount of oxygen dissolved in water. Dissolved oxygen is either from the atmosphere or released by plants during photosynthesis in water. Its concentration is between 5 to 8 mg/dm³. The minimum level of DO should be 5 mg/dm³. Below 5 mg/dm³, the water is supposed to be polluted.

Dissolved oxygen (DO) and biochemical oxygen demand (BOD) have an inverse relationship. As bacteria consume oxygen in the water while decomposing organic matter, BOD increases, leading to lower dissolved oxygen (DO) level.

Winkler method: Winkler method is a technique used to measure the amount of dissolved oxygen (DO) in water samples. It is based on the idea that dissolved oxygen (DO) in the sample solution has oxidizing properties and reacts with KI. The amount of iodine formed is measured by titrating with sodium thiosulfate using starch as an indicator which turns the solution dark blue, indicating the endpoint. The amount of iodine formed is proportional to the amount of dissolved oxygen in the solution.