Electrochemistry PDF

# Electrochemistry ## Section I 1. Electrochemistry is a branch of physical chemistry which deals with the study of interconversion between electrical energy to chemical energy. - Electrical energy → Electricity - Chemical energy → Redox Reaction 2. Redox is a simultaneous process of o...

# Electrochemistry ## Section I 1. Electrochemistry is a branch of physical chemistry which deals with the study of interconversion between electrical energy to chemical energy. - Electrical energy → Electricity - Chemical energy → Redox Reaction 2. Redox is a simultaneous process of oxidation and reduction. - **Oxidation** is the loss of e- - No. of electropositive increases - te- RHS - Reducing agent [Reductant] - **Reduction** is the gain of e- - No. of electronegative decreases - te- LHS - Oxidizing agent [oxidant] - Reduction is gain of e- - OIL RIG - Oxidation is loss of e- 3. Redox - **Spontaneous** process by its own [CE → EE] - Generate - **Non-spontaneous** process, cannot take place on its own [EE → CE] - Supply 4. **Electrolyte:** A substance which gets dissociated into positive ions (cation) and negative ions (anion) is called electrolyte. - If a substance is completely dissociated it is called a strong electrolyte [NaCl, HCl]. - If a substance is partially dissociated it is called a weak electrolyte [CH₂COOH, NH₄OH]. - A substance which cannot dissociate itself into ions is called as non-electrolyte [ Sugar, Urea] 5. **Molten NaCl vs. Aqueous NaCl:** - Molten NaCl is a liquid form of NaCl, and is a strong electrolyte. - Aqueous NaCl is a solution of NaCl in water, and is also a strong electrolyte. 6. **Electrolysis:** A process of breaking ionic compound into cation and anion by passage of electrical energy is called electrolysis. - A device in which electrolysis process takes place is called an electrolytic cell. 7. **Electrolytic Cell:** [Electrolysis of Fused / Molten NaCl] - Graphite rod (anode) → Graphite Rod (cathode) - Fused NaCl (electrolyte) - Iron vessel - Battery (e- pump) - Construction: It consists of an iron vessel in which two graphite rods are dipped in fused NaCl, which acts as an electrolyte. - A graphite rod connected to the positive terminal of a battery acts as an anode (+ve), while the graphite rod connected to the negative terminal of a battery acts as a cathode (-ve). - When electricity is passed through NaCl, it gets dissociated into Na+ and Cl-. - **Oxidation half:** 2Cl- → Cl₂ + 2e- (anion) - Cl-(anion) moves towards anode and gets discharged due to oxidation - **Reduction half:** 2Na+ + 2e- → 2Na (s) (cation) - Na+ (cation) moves towards cathode and gets discharged due to reduction. - **Net reaction:** 2Na+ + 2Cl- → 2Na + Cl₂ (g) - [Redox] - Due to electrolysis of NaCl, sodium gets deposited at the cathode and chlorine gas gets liberated to the anode. 8. **Electricity/ Farady:** - **Flow of e- **OR **Flow of charge** - Units: Coulomb (Q) or Farady (F). - **Current**: - Rate of flow of e- per unit time - I = Flow e- / time - **Voltage**: - push of e- / potential difference / EMF 9. **Coulomb:** (Q or C) - It is a quantity of electricity generated when 1 Ampere current pass for 1 second . - I.e., Q = It - Where, I = electricity, I = current, (Ampere), t = time (second) 10. **Farady (F)** - It is a quantity of electricity supplied to deposit or liberate 1 gram equivalent of substance. - I.e., F = 1 gram equivalent / molar mass / valency = 1 gram equivalent - = molar mass / valency - = 1 mole of e- - = 6.022 x 1023 no. of e- [NA] 11. **1F = 96500 coulomb** 12. **Calculate the amount of Zn deposited at the cathode when 2 F electricity is passed through ZnCl₂ solution.** - [molar mass]Zn = 64 gmol-1. - Ans: 1 F = MM / val = 64 / 2 = 32 g - OR: Zn2+ + 2e- → Zn - 2 F = 64 g 13. **Calculate the amount of electricity in F and C to deposit 27 g of Al from AICI₃ solution.** - [MM of Al] = 27 gmol-1 - Ans: 1 F = MM / val = 27 / 3 = 9 gram - OR: Al3+ + 3e- → Al - 3F = 27 g - X = 27 = 3F = 3 x 96500 C 14. **Calculate the amount of electricity in F and C for reduction of Cr3+ to Cr** - Cr3+ + 3e- → Cr [Half Reduction] - Ans: for 3 moles, 3 Cr3+ + 9e- → 3 Cr - We know, 1 F = 96500 C = 1 mole of e- - 9F or 9 x 96500 C electricity is required for reduction of 3 moles of Cr3+ to Cr. 15. **Calculate the amount of electricity required to deposit 1 mol of Cl₂ from AlCl₃ soln** - AICI₃ → 3Cl₂ + 3e- - Al³+ 3Cl- → 2Cl- + Cl₂ + 2e- - Ans: We know, 1 F = 96500 C = 1 mole of e- - 2F or 2 x 96500 C electricity is required to deposit 1 mol of Cl₂ from AlCl₃ soln. 16. **Faraday's I law of electrolysis & its quantitative aspect or analysis:** - **Statement:** Weight or mass of a substance deposited or liberated at the electrode during electrolysis is directly proportional to the quantity of electricity. - I.e., w α Q, where, Q = It. - **Quantitative aspect or analysis:** - Moles of e- = Q / 96500 = It /96500 - Mole ratio = moles of product / moles of e- - Moles of product = mole ratio x Q /96500 = mole ratio x It/96500 - Mass of product = Q / 96500 x mole ratio x molar mass = It/96500 x mole ratio x molar mass 17. **Faraday's II law of electrolysis & its quantitative aspect or analysis:** - **Statement:** When the same amount of electricity passes through 2 different electrolytes connected in series, then the weight or mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to its equivalent weight. - I.e., W α E, where, E = equivalent weight = molar mass / valency = chemical equivalent - **Quantitative aspect or analysis:** - [mass]A = [mole ratio x molar mass]A / [mole ratio x molar mass]B = [mass]B 18. **Calculate the amount of copper deposited at the cathode when 1.5 A current is passed through CuCl₂ solution for 30 mins. Molar mass of copper = 63.5 gmol-1** - **Ans:** Half reaction: Cu2+ + 2e- → Cu - [reduction half] - Mole ratio = moles of Cu / moles of e- = 1/2 - According to the quantitative analysis of Faraday's I law, [mass]Cu = Q / 96500 x mole ratio x molar mass = 1.5 x 1800 x 1/2 x 63.5 / 96500 = 13.5 x 63.5 / 965 = AL [1.1303 + 1.8028 - 2.9845] = AL [1.9486] = 8.884 x 10-1 = 0.8884 g 19. **Calculate the amount of Al deposited at the cathode when 1988 C electricity is passed through AlCl₃ soln. [Molar mass] Al = 27 gmol-1** - **Ans:** Given: Q = 1988 C - [Molar mass] Al = 27 gmol-1 - To find: [mass] Al - Half reaction: Al³+ + 3e- → Al - [reduction half] - Mole ratio = moles of Al / moles of e- = 1/3 - According to the quantitative analysis of Faraday's I law of electrolysis, [mass] AI = Q / 96500 x mole ratio x molar mass = 1988 / 96500 x 1 / 3 x 27 = 1998 x 9 / 96500 = Al [2.2985 + 0.9542 - 2.3243 - 4.9845] = AL [1.855 x 10-1] = 0.1855 g. 20. **Calculate the time required to deposit 0.7818 g of Zn from ZnCl₂ solution when 2 A current is passed through it. [Molar mass]Zn = 64 gmol-1** - **Ans:** Given: [mass]Zn = 0.7818 g, I = 2A, [molar mass]Zn = 64 g mol-1 - To find: Time (t) - Half Reaction: Zn2+ + 2e- → Zn - [reduction half] - Mole ratio = moles of Zn / moles of e- = 1 / 2 - According to the quantitative aspect of Faraday's I law, [mass]Zn = It / 96500 x mole ratio x molar mass = 0.7818 = 2 x t x 1 / 2 x 64 / 96500 - t = 0.7818 x 96500 / 64 = Al [7.8931 + 4.9845 - 4.8776 - 1.8062] = AL [3.0714] = 1.179 x 10³ = 1179 seconds 21. **Calculate current strength to deposit 1.88 g of Al when current is passed through AlCl₃ soln for 30 minutes. [Molar mass] Al = 27 gmol-1** - **Ans:** Given: [mass]Al = 1.88 g, t = 30x60 = 1800 s, [molar mass]Al = 27 gmol - To find: I. - Half reaction: Al³+ + 3e- → Al - [reduction half] - Mole ratio = moles of Al / moles of e- = 1 / 3 - According to the quantitative aspect of Faraday's I law, [mass]AI = It / 96500 x mole ratio x molar mass = 1.88 = I x 1800 x 1/ 3 x 27 / 96500 - I = 1.88 x 96500 / 1800 x 1/3 x 27 = AL [0.2742 + 2.9845 - 3.2587 + 2.2095 - 1.0492] = AL [1.0492] = 1.112 x 10¹ = 11.12 A 22. **Same amount of electricity is passed through ZnSO₄ and FeSO₄ soln respectively. 0.5919 g of Zn is deposited. Calculate amount of Fe deposited. [molar mass]Zn = 64 g, [molar mass]Fe = 56 gmol-1** - **Ins:** To find: [mass]Fe - Half reactions: - Fe²+ + 2e- → Fe mole ratio = 1/2 - Zn²+ + 2e- → Zn mole ratio = 1/2 - According to quantitative aspect of Faraday's II law, [mass]Zn = [mole ratio x molar mass]Zn / [mole ratio x molar mass]Fe = [mass]Fe - [mass]Fe = 0.5919 x 64 / 56 = AL [1.7723 - 1.7482 + 1.5205 - 1. 8062] = AL [1.7143] = 5.18 x 10-1 = 0. 518 g 23. **Calculate the amount of Mg deposited & Cl₂ liberated when 2.7 A current is passed through MgCl₂ solution, for 1 hour. [molar mass]Mg = 24 gmol-1; [molar mass]Cl₂ = 71 gmol-1** - **Ans:** [Molar mass]Mg = 24 g mol-1; [Molar maiss]Cl₂ = 71 gmol-1 - I = 2.7 A, t = 1 hr = 3600 S - **Half reaction:** Mg 2+ + 2e- → Mg - [reduction] - **MOLE ratio:** moles of product / moles of electron = 1 / 2 - **[mass]Mg:** = It / 96500 x mole ratio x molar mass = 2.7 x 3600 x 1 / 2 x 24 / 96500 = 2.7 x 36 x 12 / 965 = AL [0.4314 + 1.5563 + 1.0792 - 2.9845] = 12.09 g - **Half reaction:** 2Cl- → Cl₂ + 2e- -[oxidation] - **Mole ratio:** moles of product / moles of electron = 1 / 2 - **[mass]Cl₂:** = It / 96500 x mole ratio x molar mass = 2.7 x 3600 x 1 / 2 x 71/ 96500 = 2.7 x 36 x 71 / 965 x 2 = AL [0.4314 + 1.8513 + 3.8390 - 2.9845 + 0.3010 - 3.2855] = 3.495 g ## Section II 1. **Conductor and its types:** - A substance which allows the flow of electricity through it is called a conductor. - **Types of conductors:** - **Electrolytic/Ionic conductor:** A substance which allows the flow of electricity through it due to the flow of cations and anions. - e.g. acid soln, salt soln - **Electronic/Metallic conductor:** A substance which allows the flow of electricity through it due to the flow of free electrons. - e.g. Al, Cu 2. **Temperature & conductance** - Temperature directly affects the **electrolytic conductance**, but has no effect on **electronic conductance** - Working of the cell is independent of temperature 3. **Ohm's Law:** - V = IR - V = Potential difference (V) - I = electric current (A) - R = Resistance - Resistance: opposition or hinderance for flow of electrons - R = V / I = potential difference (V) / electric current (A) - 1 Ω (omega) = IVA-1 - 1 ohm = IVA-1 4. **Conductance (G):** - G = 1 / R = reciprocal of resistance - G = I / V = electric current (A)/potential difference (V) - Conductance is the ease with which electrons flow - 1 Siemens (S) = 1 ohm-1 = 1 Av-1 5. **Cell Constant (b) or G**: - It is the ratio of distance between two electrodes in cm (l) to the cross-sectional [C/S] area of electrode in cm² (α) - I.e., b = l / α = cm / cm² = cm-1 - Where, l = distance between 2 electrodes (cm), α = C/S area of electrode (cm²) 6. **Resistivity or Specific Resistance (९) or Rho ** - R α l / a - R α b - R = ९b - ९ = R / b = Resistance / Cell constant - ९ = ohm = ohm.cm / cm-1 - **Resistivity (९)** is the ratio of resistance (R) to the cell constant (b). - Resistivity (९) is resistance (R) when the cell constant (b) is unity [b=1] - Resistivity (९) is resistance (R) when the distance between 2 electrodes (l) is 1 cm and the C/S area of the electrode (a) is 1 cm². 7. **Conductivity or Specific Conductance (K) or Kappa** - G α 1 / R - G α b - G = Kb - K = Gb / b = R / ९ = 1 / ९ - K = ohm² cm²¹ = ohm² cm²¹ / R = Scm-1. - **Conductivity (K)** is the product of the conductance (G) and the cell constant (b). - It is the ratio of cell constant (b) to resistance (R). - It is the reciprocal of resistivity (९). - Conductivity (K) is conductance (G) when the cell constant is unity [b=1] - Conductivity (K) is conductance (G) when the distance between the 2 electrodes (l) is 1 cm and the C/S area of the electrode (a) is 1 cm2. 8. **Molar Conductivity (^):** - **Definition:** The conductance of all ions produced due to ionization of 1 mole of electrolyte in 'V' mL of the solution. - The conductance of all ions produced due to ionization of 1 mole of electrolyte in 1000 mL of the solution. - **OR:** The ratio of conductivity (K) to molar concentration - ^ = 1000K / C - Where: K = conductivity = ohm-cm-1, C = molar concentration = mol/cm³ - **Unit of ^:** - ohm−1 cm−1 / [mol] / cm³ = ohm-1 mol-1 cm² = Smol-1 cm² = 1000K / C = AC = 1000K - **SI unit of ^:** ohm mol-m² OR ^ = Smol m² - ^ = 1000K / C - **Relationship between ^ & dilution:** ^ α 1 / dilution - ^ α K x dilution , K α con. x 1/ dilution, K α 1/ dilution 9. **Kohlrausch equation:** - It is a graphical method to determine ^o (molar conductivity at zero concentration) or ^∞ (molar conductivity at infinite dilution). - Kohlrausch equation: Λ = -avC + ^∞ - Λ = molar conductivity of the electrolyte - a = Kohlrausch constant - c = molar concentration. - ^∞ = molar conductivity at zero concentration - There is no increase in the value of ^ when the solution is so diluted that further dilution is not possible. - Graphical representation of the equation: plot ^ on the y-axis and √C on the x-axis - y-intercept = ^∞ - Slope = a (Kohlrausch constant) - **Strong electrolyte**: Linear relationship between Λ and √C, which gives the value of ^∞ graphically. - **Weak electrolyte:** Non-linear relationship between Λ and √C, which doesn't give the value of ^∞ graphically. 10. **Kohlrausch law of independent migration of ions and its application:** - **Statement:** Molar conductivity of the electrolyte at zero concentration is the summation of molar conductivity of positive ion (cation) and the molar conductivity of negative ion (anion) at zero concentration. - I.e., ^∞ = n+^∞ + n- ^∞ - Where, ^∞ = molar conductivity of the electrolyte at zero concentration. - n = no. of ions - ^∞ = molar conductivity of ions at zero concentration - e.g., ^ ∞ = ^ ∞ + 3^∞ = 2^ ∞ + 3^ ∞ - AICI_3 - Al³⁺ 3Cl⁻ - **Application:** Kohlrausch law is used to determine ^∞ for the electrolyte when ^o of ions are known. 11. **Degree of Dissociation (X):** - The degree of dissociation (X) is the ratio of molar conductivity (Λ) to molar conductivity at zero concentration (^∞). - I.e., X = Λ / ^∞ - The value of X is always in the range of 0 to 1. 12. **Calculate the molar conductivity of 0.035 M AgNO₃ soln when its conductivity is 3.228 x 10⁻³ Scm⁻¹. ** - **Ans:** Given: c = 0.035 M = 35 x 10⁻³ M, K = 3.228 x 10⁻³ Scm⁻¹ - Λ = 1000K / C = 1000 x 3.228 x 10⁻³ / 35 x 10⁻³ = 3228 / 35 = AL [3.5090 - 1.5441 + 1.9649] = 92.23 Smol-1cm² 13. **What is ^∞ for HCl and CH3COOH: ** - **Ans:** - (a) 80 80 - (d) NA 80 14. **Daniel cell:** - It consists of 2 half cells: - **Zn half cell (oxidation half):** - Zn rod dipped in ZnSO₄ soln, where Zn rod acts as the anode (-). - **Cu half cell (reduction half):** - Cu rod dipped in CuSO₄ soln, where Cu rod acts as cathode (+). - The two half cells are connected externally by a metallic wire and internally by a salt bridge. - **Salt bridge:** allows migration of ions. - **Working:** - At anode, oxidation takes place: - Zn → Zn2+ + 2e- - Zn atoms from the Zn rod are oxidized to give Zn2+ ions which pass into the solution while the electrons will remain on Zn rod (anode). - Anode is negatively charged - At the cathode, reduction takes place: - Cu2+ + 2e- → Cu - Electrons are transferred from the anode to cathode via an external circuit (metallic wire) and gained by the Cu2+ ion. - Net reaction: - Zn + Cu 2+ → Zn 2+ + Cu - Redox - Cell Formula / Notation: - Zn | ZnSO4 (C₁) || CuSO4 (C₂) | Cu - LOAN = left Oxi Anode, Zn | Zn2+ (C₁) || Cu2+ (C₂) | Cu - **Conventions/Rules for writing cell formula or cell notation:** - Zn | Zn2+ (C₁) || Cu2+ (C₂) | Cu - - left Oxi Anode, + right red Cathode 15. **Lead accumulator cell (galvanic cell) - discharging:** - Construction: - Acid proof vessel is filled with 38% H₂SO₄ soln, which acts as an electrolyte. - Series of Pb plates act as the anode, and series of Pb plates coated with PbO₂ act as the cathode. - **Working:** - **Discharging - galvanic cell**: - At the anode: - Pb → Pb2+ + 2e- - Pb atoms from the Pb plate are oxidized to give Pb2+ ions which combine with SO₄²- ion from the H₂SO₄ solution to form insoluble PbSO₄. - Anode is negatively charged. - At the cathode: - PbO2 (IV) + 4H+ + SO₄²- + 2e- → PbSO₄ + 2H₂O - Pb(IV)O₂ gets reduced to give Pb2+ and then Pb2+ combines with SO₄²- from H₂SO₄ solution to form an insoluble PbSO₄ along with water. - **Charging - electrolytic cell:** - The reactions are reversed. - Net reaction during charging: 2PbSO4 + 2H2O → Pb + PbO₂ + 2H₂SO₄ - [redox] - Electrolysis of PbSO₄ takes place with Pb deposited at the anode, PbO₂ at the cathode, and H₂SO₄ regenerated. - The concentration of H₂SO₄ increases from 1.17 g/ml to 1.20 g/ml, and the cell gets charged. 16. **H₂-O₂ Fuel Cell:** - H₂ gas acts as fuel, and O₂ acts as oxidizer. - The chemical reaction between fuel and oxidizer produces water as a by-product and electrical energy as a main product. - **Construction:** - It consists of a pair of porous carbon electrodes, one acts as anode and another acts as cathode. - H₂ gas is passed through the anode compartment and O₂ gas is passed through the cathode compartment. - Anode and cathode are separated by KOH or NaOH sol" which acts as an electrolyte. - **Working:** - At the anode, **oxidation** takes place: - 2H₂ → 4H+ + 4e- - H₂ gas is oxidized to give H+ ions which combine with OH- ions from KOH soln to form H₂O. - At the cathode, **reduction** takes place: - 2H₂O + O₂ + 4e- → 4OH - (ORR) - Dissolved oxygen reduced to give OH- which maintains a constant concentration of KOH throughout working. - Net reaction: - 2H₂ + O₂ → 2H₂O + E.E. (1.23V) -[redox] - **Advantages/ Applications:** - It is used in the Apollo Space Program, for a pollution-free E.E. - It produces pollution-free E.E,. The efficiency of the cell is about 70%, which is more than conventional power plants. - It can be used at remote locations such as war and space. - It produces water as a by-product which can be used for drinking purposes for astronauts. 17. **EMF series and its applications:** - **EMF series**: - Electromotive series - Electrochemical series - Activity series - Reactivity series. - **E° = SRP** - F₂ : +2.87 V - High SRP → reduction → cathode - Cu: +0.34 V - H₂: 0 V - Zn: -0.76 V - Li: -3.05 V - high SOP → oxidation→ anode - **Elements in the series are arranged in decreasing order of their SRP value.** - It is also be called an **electrochemical/electrometric or activity or reactivity series.** - **Applications:** - It is used to detect anode and cathode, and predict which redox reaction will spontaneously occur. - Elements with a **higher SRP** quickly undergo reduction and act as the **cathode**, while elements with a **lower SRP** quickly undergo oxidation and they are the **anode** - **Selection of good oxidizing agents**: - Elements with a **high SRP** quickly undergo reduction, making them **good oxidizing agents**. - **Selection of good reducing agents:** - Elements placed at the lower part of the series with a **high SOP** undergo oxidation and are **good reducing agents**. - It is used to determine the **standard EMF of the cell** (E°cell): - E°cell = E°cathode - E°anode - E° = standard potential - E° = SRP - It is used to determine the **spontaneity of the redox reaction** - If E°cell is positive, the redox reaction is spontaneous. - If E°cell is negative, the redox reaction is non-spontaneous. 18. **Electrolysis of aqueous NaCl:** - The electrolyte is the salt solution. - **At the anode:** 2Cl⁻ → Cl₂ (g) + 2e- - oxidation half - **At the cathode:** 2H+ + 2e- → H₂(g) - reduction half - **Net reaction:** 2H⁺ + 2Cl⁻ → H₂↑ + Cl₂↑ - redox reaction - NaOH remains in the solution. 19. **How to determine the cell constant (b) using a Wheatstone bridge circuit?** - **Wheatstone Bridge circuit** is used to determine unknown resistance. - Construction: The Wheatstone bridge circuit is made up of platinum or copper wire connected to two resistances - an unknown (R) and a known (Rs) - through a detector (D) and jockey/slider/wiper. The Wheatstone bridge circuit is also connected to the cell having a known conductivity (K) and an unknown cell constant (b). - **Working:** - A slider or jockey is slid from point B to point C in such a way that the detector shows zero deflection or no deflection. This means the circuit is in a balanced condition. - **According to the Wheatstone bridge principle:** R / (BM) = Rs / (CM) - **Determine the resistance of the cell:** - We know, b = K x Rcell 20. **Ni-Cd cell:** - **Definition:** Ni-Cd cell is a secondary (2°) dry cell because it generates as well as stores electrical energy. - **Construction:** - **Anode:** Cadmium metal - **Cathode:** Nickel (IV) oxide, NiO₂, supported on Ni - It uses basic KOH as the electrolyte. - **Working:** - **At the anode:** Cd + 2OH⁻ → Cd(OH)₂ + 2e- - oxidation - **At the cathode:** NiO₂ + 2H₂O + 2e- → Ni(OH)₂ + 2OH- - reduction - Ni⁴⁺ ions are reduced to Ni²⁺ ions which further gives Ni(OH)₂ - **Net reaction:** Cd + NiO₂ + 2H₂O → Cd(OH)₂ + Ni(OH)₂ - **Applications:** - The potential of the cell is about 1.4 V. - The cell has a longer life than other dry cells - It can be used in electronic watches, calculators, photographic experiments, etc. 21. **Standard hydrogen electrode (SHE):** - **Definition:** It is a reference electrode. The potential value is arbitrarily or assumed to be fixed or constant. - **Types of reference electrodes:** - **Primary (1°) reference electrode**: SHE - **Secondary (2°) reference electrode**: calomel (Hg₂Cl₂) - **Construction:** - The SHE is made of glass: Outer glass jacket with an inlet for pure and dry H₂ gas (1 atm), which is sealed with a small amount of Hg (mercury). Inside the glass jacket, there is a glass tube which contains a Pt/Cu wire which is sealed with a small amount of mercury. This is further connected to the Pt plate, and the upper part of the wire is connected to the electrode. The whole assembly is in 1 M HCl soln (soln of H+ ions at unit concentration). - **Working:** - When H₂ gas is passed through the glass jacket, it is adsorbed by the Pt plate and undergoes either oxidation or reduction. - **At the anode:** H₂ + 2H+ + 2e- - oxidation - **At the cathode:** 2H+ + 2e- → H₂ - reduction - The extent of the reaction is very small, which means the potential developed is also very small, which is assumed to be 0. E°SHE = 0. - **Representation:** Pt, H₂(1 atm) | H+ (1M) - - **Applications:** - It is used to determine the unknown potential of electrodes. - e.g., Zn / Zn²⁺(0.1M) || H⁺(1M) | H₂(1atm), Pt. 22. **Write the cell formula, cell reaction and calculate the standard EMF of the cell for Al | Al³⁺ & SHE when E°Au = +1.78 V** - **Ans:** Cell Formula: Pt, H₂(1atm) | H+ (1M) || Au³⁺ (1M) | Au - **Cell reaction:** - **At the anode:** - H₂(g) → 2H+(aq) + 2e- - oxidation - **At the cathode:** - 3H₂(g) → 6H+(aq) + 6e- - Au³⁺(aq) + 3e- → Au(s) - reduction - **Net reaction:** 2Al + 3Cu2+ → 2Al3+ + 3Cu - Redox