Electrochemistry

1. Electrochemical Cells:

• Galvanic Cells (Voltaic Cells): Devices that convert chemical energy into electrical energy through redox reactions. Key components include two half-cells (anode and cathode), a salt bridge, and an external circuit.Electrolytic Cells: Devices that use electrical energy to drive non-spontaneous redox reactions (electrolysis). They also consist of an anode and cathode, but the setup and purpose are different from galvanic cells.

2. Electrode Potentials:

• Standard Electrode Potential (E°): The potential of a half-cell under standard conditions (298 K, 1 M concentration, 1 atm pressure). It’s a measure of the tendency of a species to be reduced.Cell Potential (E°cell): The difference in standard electrode potentials between the cathode and anode. E°cell = E°(cathode) – E°(anode). Nernst Equation: Relates electrode potential to non-standard conditions (different concentrations, temperatures). It allows calculation of cell potentials under varying conditions.

3. Thermodynamics of Electrochemical Cells:

• Gibbs Free Energy (ΔG): Related to cell potential by ΔG = -nFE°cell, where n is the number of moles of electrons transferred and F is Faraday’s constant. A negative ΔG indicates a spontaneous reaction.Equilibrium Constant (K): Related to standard cell potential by ΔG° = -RTlnK. This connects electrochemistry to chemical equilibrium.

4. Electrolysis:

• The process of using electrical energy to drive non-spontaneous chemical reactions.Faraday’s Laws of Electrolysis: Quantify the relationship between the amount of charge passed through an electrolytic cell and the amount of substance deposited or liberated at the electrodes.

5. Batteries:

• Primary Batteries: Non-rechargeable batteries (e.g., dry cell, mercury cell).Secondary Batteries: Rechargeable batteries (e.g., lead-acid battery, nickel-cadmium battery, lithium-ion battery).Fuel Cells: Devices that convert the chemical energy of a fuel (e.g., hydrogen) and an oxidant (e.g., oxygen) directly into electricity.

6. Corrosion:

• The electrochemical process of deterioration of metals due to reaction with the environment.Factors affecting corrosion and methods for its prevention (e.g., galvanization, cathodic protection).

7. Conductivity:

• Metallic Conductance: Electron flow through metals.Electrolytic Conductance: Ion flow through electrolyte solutions.Molar Conductivity (Λm): Conductivity of a solution containing 1 mole of electrolyte.Kohlrausch’s Law: The molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar conductivities of its constituent ions.

Exercise1.  Arrange the following metals in the order in which they displace each other from their salts.Al, Cu, Fe, Mg and ZnAns : Mg (Magnesium)Al (Aluminum)Zn (Zinc)Fe (Iron)Cu (Copper)2. Given the standard electrode potentials, K+/K=-2. 93 V, Ag+/Ag = 0.80 V, Hg2+/Hg =0.79V, Mg2+/Mg=-2.37V, Cr3+/Cr=0.74V.Arrange these metals in their increasing order of reducing power.Ans : 

1. Ag (Silver): Ag⁺/Ag = +0.80 V (Most positive E°, weakest reducing agent)Hg (Mercury): Hg²⁺/Hg = +0.79 VCr (Chromium): Cr³⁺/Cr = -0.74 V  Mg (Magnesium): Mg²⁺/Mg = -2.37 V  K (Potassium): K⁺/K = -2.93 V (Most negative E°, strongest reducing agent)

So, the order is: Ag < Hg < Cr < Mg < K3.  Depict the galvanic cell in which the reactionZn(s) + 2Ag+(aq) —-> 7M2+(aq) + 2Ag (s) takes place. Further show:(i) Which of the electrode is negatively charged?(ii) The carriers of the current in the cell.(iii) Individual reaction at each electrode.Ans : Galvanic Cell Depiction:Zn(s) | Zn²⁺(aq, 1M) || Ag⁺(aq, 1M) | Ag(sExplanation of the Depiction:

• Zn(s) | Zn²⁺(aq, 1M): This represents the anode half-cell. Solid zinc (Zn) is in contact with a 1M solution of zinc ions (Zn²⁺). ||: This double vertical line represents the salt bridge. Ag⁺(aq, 1M) | Ag(s): This represents the cathode half-cell. Silver ions (Ag⁺) in a 1M solution are in contact with solid silver (Ag).

(i) Negatively Charged Electrode:The anode is negatively charged. This is because oxidation (loss of electrons) occurs at the anode, releasing electrons that give it a negative charge.(ii) Carriers of Current:

• In the external circuit: Electrons flow from the negatively charged anode (Zn) to the positively charged cathode (Ag).In the salt bridge: Ions (cations and anions) carry the current. Cations move toward the cathode half-cell, and anions move toward the anode half-cell, balancing the charge buildup due to the electron flow.In the solutions: Ions (Zn²⁺ and Ag⁺, along with the counter-ions from the salts used) are charge carriers.

(iii) Individual Reactions at Each Electrode:

• Anode (Oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻Cathode (Reduction): 2Ag⁺(aq) + 2e⁻ → 2Ag(s)

4. Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.Ans : 5. Write the Nernst equation and emf of the following cells at 298 K:Ans : (i) Mg(s) | Mg²⁺(0.001 M) || Cu²⁺(0.0001 M) | Cu(s)

1. Standard Cell Potential (E°_cell):

2. Anode (Oxidation): Mg → Mg²⁺ + 2e⁻ (E° = +2.37 V, sign changed for oxidation)Cathode (Reduction): Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)

E°_cell = (+0.34 V) – (+2.37 V) = 2.71 VMoles of Electrons Transferred (n): n = 2Reaction Quotient (Q): Q = [Mg²⁺] / [Cu²⁺] = (0.001 M) / (0.0001 M) = 10Nernst Equation: E_cell = 2.71 V – (0.0592 V / 2) * log(10) E_cell = 2.71 V – 0.0296 V = 2.68 V

(ii) Fe(s) | Fe²⁺(0.001 M) || H⁺(1 M) | H₂(g) (1 bar) | Pt(s)

1. Standard Cell Potential (E°_cell):

2. Anode (Oxidation): Fe → Fe²⁺ + 2e⁻ (E° = +0.44 V)Cathode (Reduction): 2H⁺ + 2e⁻ → H₂ (E° = 0.00 V)

E°_cell = (0.00 V) – (+0.44 V) = -0.44 VMoles of Electrons Transferred (n): n = 2Reaction Quotient (Q): Q = [Fe²⁺] / [H⁺]² = (0.001 M) / (1 M)² = 0.001Nernst Equation: E_cell = -0.44 V – (0.0592 V / 2) * log(0.001) E_cell = -0.44 V – (-0.0888 V) = -0.351 V

(iii) Sn(s) | Sn²⁺(0.050 M) || H⁺(0.020 M) | H₂(g) (1 bar) | Pt(s)

1. Standard Cell Potential (E°_cell):

2. Anode (Oxidation): Sn → Sn²⁺ + 2e⁻ (E° = +0.14 V)Cathode (Reduction): 2H⁺ + 2e⁻ → H₂ (E° = 0.00 V)

E°_cell = (0.00 V) – (+0.14 V) = -0.14 VMoles of Electrons Transferred (n): n = 2Reaction Quotient (Q): Q = [Sn²⁺] / [H⁺]² = (0.050 M) / (0.020 M)² = 125Nernst Equation: E_cell = -0.14 V – (0.0592 V / 2) * log(125) E_cell = -0.14 V – 0.109 V = -0.249 V

(iv) Pt(s) | Br₂(l) | Br⁻(0.010 M) || H⁺(0.030 M) | H₂(g) (1 bar) | Pt(s)1. Identify the Half-Reactions and Standard Cell Potential (E°_cell):

• Anode (Oxidation): 2Br⁻(aq) → Br₂(l) + 2e⁻

• E°(Br₂/Br⁻) = +1.07 V (You’ll need to look this up in a standard reduction potential table)E°(Br⁻/Br₂) = -1.07 V (Sign changed for oxidation)

Cathode (Reduction): 2H⁺(aq) + 2e⁻ → H₂(g)

• E°(H⁺/H₂) = 0.00 V (By definition)

E°_cell = E°(Cathode) – E°(Anode) E°_cell = (0.00 V) – (-1.07 V) = +1.07 V

2. Determine the Number of Electrons Transferred (n):

• n = 2 (2 electrons are transferred in the balanced redox reaction)

3. Calculate the Reaction Quotient (Q):

• Q = ([H⁺]² / [Br⁻]²)Q = (0.030)² / (0.010)² = 9

4. Apply the Nernst Equation (Simplified at 298 K):E_cell = E°_cell – (0.0592 V / n) * log(Q)E_cell = 1.07 V – (0.0592 V / 2) * log(9)E_cell = 1.07 V – (0.0296 V) * (0.954)E_cell = 1.07 V – 0.0283 VE_cell ≈ 1.042 V7.  In the button cells widely used in watches and other devices the following reaction takes place:Ans : 8. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.Ans : Conductivity (κ)

• Definition: Conductivity (represented by the Greek letter kappa, κ) measures a solution’s ability to conduct electricity. It’s the reciprocal of resistivity. Essentially, it tells you how easily charges can move through the solution.  Units: Siemens per meter (S/m) or ohm⁻¹ meter⁻¹ (Ω⁻¹m⁻¹)  Factors Affecting Conductivity:

• Number of ions: More ions mean higher conductivity.  Charge of ions: Higher charge means greater attraction/repulsion, affecting ion mobility.Mobility of ions: How easily ions move through the solution (related to size, shape, and interactions with solvent).Concentration of ions: Generally, increasing concentration increases conductivity (up to a point).Temperature: Higher temperature usually increases conductivity.  

Molar Conductivity (Λm)

• Definition: Molar conductivity is the conductivity of a solution containing one mole of the electrolyte dissolved in a specific volume of the solution. It essentially normalizes conductivity with respect to the amount of dissolved electrolyte.  Units: Siemens meter squared per mole (S m²/mol) or ohm⁻¹ meter² mol⁻¹ (Ω⁻¹ m² mol⁻¹)  Formula: Λm = κ / c

• where κ is the conductivity and c is the molar concentration of the electrolyte.  

Variation with ConcentrationThe way conductivity and molar conductivity change with concentration is different, and it’s essential to understand the distinction:

• Conductivity (κ):

• Generally increases with increasing concentration because there are more ions available to carry charge.  However, at very high concentrations, conductivity might decrease slightly due to interionic interactions hindering ion movement.

Molar Conductivity (Λm):

• Decreases with increasing concentration for strong electrolytes. This is because the increase in the number of ions is less proportional to the increase in concentration due to increased interionic interactions which restrict the movement of ions.Increases with decreasing concentration (and dramatically so at very low concentrations) for weak electrolytes. Weak electrolytes dissociate only partially into ions in solution. As you dilute the solution (decrease concentration), the degree of dissociation increases, leading to a larger increase in the number of charge-carrying ions than in the case of strong electrolytes. This leads to a higher molar conductivity at lower concentrations for weak electrolytes.

8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.Ans : 9. The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 x 10-3 S cm-1?Ans : 1. Understand the relationship:The relationship between resistance (R), conductivity (κ), and the cell constant (G*) is given by:R = G*/κ2. Rearrange the formula to solve for the cell constant (G):*G* = R * κ3. Plug in the given values:

• R = 1500 Ω  κ = 0.146 x 10⁻³ S cm⁻¹  

G* = 1500 Ω * 0.146 x 10⁻³ S cm⁻¹ = 0.219 cm⁻¹  4. Convert the cell constant to m⁻¹ (if required)Since 1 m = 100 cm, 1 cm = 0.01 m and 1 cm⁻¹ = 100 m⁻¹  G* = 0.219 cm⁻¹ * 100 cm/m = 21.9 m⁻¹Therefore, the cell constant is 0.219 cm⁻¹ or 21.9 m⁻¹10. The conductivity of NaCl at 298 K has been determined at different concentrations and the results are given below:Ans : 11. Conductivity of 0.00241 M acetic acid is 7.896 x 10-5 S cm-1. Calculate its molar conductivity. If Λm0, for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?Ans : 1. Calculate the molar conductivity (Λm):

• Convert conductivity to S/m: Conductivity is given in S/cm, but we need it in S/m for the molar conductivity calculation. 1 S/cm = 100 S/m So, 7.896 x 10⁻⁵ S/cm = 7.896 x 10⁻³ S/m  Use the formula: Λm = κ / c Where: * κ = conductivity = 7.896 x 10⁻³ S/m * c = concentration = 0.00241 M = 0.00241 mol/L  Λm = (7.896 x 10⁻³ S/m) / (0.00241 mol/L) Λm ≈ 3.276 S m²/mol

2. Calculate the degree of dissociation (α):

• Use the formula: α = Λm / Λm⁰ Where:

• Λm = molar conductivity at given concentration ≈ 3.276 S m²/molΛm⁰ = molar conductivity at infinite dilution = 390.5 S cm²/mol = 390.5 x 10⁻⁴ S m²/mol = 0.03905 S m²/mol

α = (3.276 S m²/mol) / (0.03905 S m²/mol) α ≈ 0.0839

3. Calculate the dissociation constant (Ka):

• Use the formula: Ka = (cα²) / (1 – α) Where: * c = concentration = 0.00241M * α = degree of dissociation ≈ 0.0839 Ka = (0.00241 M * (0.0839)²) / (1 – 0.0839) Ka ≈ 1.86 x 10⁻⁵

12.  How much charge is required for the following reductions:(i) 1 mol of Al3+ to Al?(ii) 1 mol of Cu2+ to Cu ?(iii) 1 mol of Mn04- to Mn2+?Ans : (i) Al³⁺ to Al:

• Moles of electrons: 3 moles  Charge: 3 moles * 96,485 C/mol = 289,455 Coulombs

(ii) Cu²⁺ to Cu:

• Moles of electrons: 2 molesCharge: 2 moles * 96,485 C/mol = 192,970 Coulombs

(iii) MnO₄⁻ to Mn²⁺:

• Moles of electrons: 5 moles  Charge: 5 moles * 96,485 C/mol = 482,425 Coulombs

13.  How much electricity in terms of Faraday is required to produce :(i) 20·0 g of Ca from molten CaCl2(ii) 40·0 g of Al from molten Al2O3 ?Ans : 14. How much electricity is required in coulomb for the oxidation of (i) 1 mol of H2O to 02 (ii) 1 mol of FeO to Fe203Ans : 15. A solution of Ni(N03)2 is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?Ans : 16. Three electrolytic cells A, B, C containing solutions of ZnS04, AgNO3 and CuS04, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 45 g of silver deposited at the cathode of call B. How long did the current flow? What mass of copper and zinc were deposited?Ans : 1. Calculate the moles of silver deposited:

• Molar mass of silver (Ag) = 107.87 g/molMoles of Ag = mass / molar mass = 45 g / 107.87 g/mol ≈ 0.417 moles

2. Determine the charge (Q) passed through the cells:

• The reduction of silver ions (Ag⁺) to silver metal (Ag) involves the transfer of 1 electron per silver ion: Ag⁺ + e⁻ → AgSince 1 mole of electrons is equivalent to 1 Faraday (96,485 Coulombs), we can calculate the charge: Charge (Q) = moles of Ag * Faraday’s constant Q = 0.417 moles * 96,485 C/mol ≈ 40,285 Coulombs

3. Calculate the time (t) the current flowed:

• Current (I) = 1.5 amperes (A)Time (t) = Charge (Q) / Current (I)t = 40,285 C / 1.5 A ≈ 26,857 seconds

4. Convert time to minutes:

• t = 26,857 seconds / 60 seconds/minute ≈ 447.6 minutes

5. Calculate the mass of copper (Cu) deposited:

• The reduction of copper ions (Cu²⁺) to copper metal (Cu) involves the transfer of 2 electrons: Cu²⁺ + 2e⁻ → CuMoles of Cu = (Charge (Q) / Faraday’s constant) / 2Moles of Cu = (40,285 C / 96,485 C/mol) / 2 ≈ 0.208 molesMass of Cu = moles of Cu * molar mass of Cu (63.55 g/mol)Mass of Cu ≈ 0.208 moles * 63.55 g/mol ≈ 13.2 g

6. Calculate the mass of zinc (Zn) deposited:

• The reduction of zinc ions (Zn²⁺) to zinc metal (Zn) involves the transfer of 2 electrons: Zn²⁺ + 2e⁻ → ZnMoles of Zn = (Charge (Q) / Faraday’s constant) / 2Moles of Zn = (40,285 C / 96,485 C/mol) / 2 ≈ 0.208 molesMass of Zn = moles of Zn * molar mass of Zn (65.38 g/mol)Mass of Zn ≈ 0.208 moles * 65.38 g/mol ≈ 13.6 g

Therefore:

• The current flowed for approximately 447.6 minutes.The mass of copper deposited is approximately 13.2 g.The mass of zinc deposited is approximately 13.6 g.

17.  Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible.(a) Fe3+(aq) and I–(aq)(b) Ag+(aq) and Cu(s)(c) Fe3+(aq) and Br–(aq)(d) Ag(s) and Fe3+(aq)(e) Br2(aq) and Fe2+(aq).Ans : 18. Predict the products of electrolysis in each of the following.(i) An aqueous solution of AgNO3 with silver electrodes.(ii) An aqueous solution of AgNO3 with platinum electrodes.(iii) A dilute solution of H2S04 with platinum electrodes.(iv) An aqueous solution of CuCl2 with platinum electrodes.Ans : (i) Aqueous AgNO₃ with Silver Electrodes:

• At the Cathode (Reduction): Silver ions (Ag⁺) will be reduced to silver metal (Ag). This is the preferred reaction because it requires less energy than reducing water to hydrogen gas. Ag⁺(aq) + e⁻ → Ag(s)  At the Anode (Oxidation): Silver metal from the electrode will be oxidized to silver ions. Ag(s) → Ag⁺(aq) + e⁻Overall: Silver metal will be deposited at the cathode, and the silver electrode at the anode will dissolve (the concentration of Ag⁺ in the solution will increase slightly).

(ii) Aqueous AgNO₃ with Platinum Electrodes:

• At the Cathode (Reduction): Silver ions (Ag⁺) will be reduced to silver metal (Ag), as in the previous case. Ag⁺(aq) + e⁻ → Ag(s)At the Anode (Oxidation): Since platinum is an inert electrode, it won’t be oxidized. Instead, water will be oxidized to oxygen gas (O₂). 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻Overall: Silver metal will be deposited at the cathode, and oxygen gas will be evolved at the anode. The solution will become more acidic due to the H⁺ ions produced.  

(iii) Dilute H₂SO₄ with Platinum Electrodes:

• At the Cathode (Reduction): Hydrogen ions (H⁺) from the sulfuric acid will be reduced to hydrogen gas (H₂). 2H⁺(aq) + 2e⁻ → H₂(g)  At the Anode (Oxidation): Water will be oxidized to oxygen gas (O₂), as platinum is inert. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻  Overall: Hydrogen gas will be evolved at the cathode, and oxygen gas will be evolved at the anode. The concentration of H₂SO₄ will increase slightly as water is consumed.

(iv) Aqueous CuCl₂ with Platinum Electrodes:

• At the Cathode (Reduction): Copper ions (Cu²⁺) will be reduced to copper metal (Cu). Cu²⁺(aq) + 2e⁻ → Cu(s)At the Anode (Oxidation): Chloride ions (Cl⁻) will be oxidized to chlorine gas (Cl₂). 2Cl⁻(aq) → Cl₂(g) + 2e⁻Overall: Copper metal will be deposited at the cathode, and chlorine gas will be evolved at the anode.