Questions: Electrochemistry

Q1 Define standard cell potential and explain how it is related to the Gibbs energy change of the cell reaction.

Answer (English): The standard cell potential is the potential of a cell when every species is in its standard state, which means unit activity for solutes, 1 bar for gases and pure solids or liquids for the condensed phases. It is the difference between the standard reduction potentials of the cathode and the anode. The Gibbs energy change of the cell reaction is related by ΔG° = -nFE°cell, where n is the number of electrons transferred and F is 96500 C mol^-1. A positive E°cell gives a negative ΔG° and a spontaneous reaction; a negative E°cell gives a positive ΔG° and a non-spontaneous reaction. The same relation holds at non-standard conditions if Ecell is used in place of E°cell.

Hinglish: Standard conditions par cell ka potential E°cell hota hai aur yeh cathode aur anode ke standard reduction potentials ka difference hai. ΔG° = -nFE°cell se energy change nikalta hai. Positive E°cell matlab reaction spontaneous aur ΔG negative; negative E°cell matlab reaction non-spontaneous.

Q2 Write the Nernst equation for the Daniell cell and calculate its potential when zinc ion is 0.1 M and copper ion is 0.01 M. E°cell is 1.10 V.

Answer (English): The cell is Zn | Zn2+ || Cu2+ | Cu, with zinc oxidised at the anode and copper deposited at the cathode. The cell reaction is Zn + Cu2+ giving Zn2+ + Cu, and two electrons are transferred. The Nernst equation is E = E° minus 0.0591 divided by 2 times log of the zinc ion concentration divided by the copper ion concentration. Substituting 0.1 and 0.01 gives Q equal to 10 and log Q equal to 1. The correction is 0.0296 V, so E is 1.10 minus 0.0296, which is about 1.07 V. The potential is lower than the standard value because the product concentration is relatively high.

Hinglish: Daniell cell mein Zn anode par oxidise hota hai aur Cu2+ cathode par deposit hota hai; n = 2. Nernst equation E = E° - (0.0591/2) log([Zn2+]/[Cu2+]). Q = 0.1/0.01 = 10, log 10 = 1, isliye correction 0.0296 V aur E ≈ 1.07 V.

Q3 Describe the standard hydrogen electrode and explain why it is assigned zero potential.

Answer (English): A standard hydrogen electrode is a platinum plate coated with finely divided platinum black. It dips into an acidic solution of hydrogen ions at unit activity, commonly 1 M, with hydrogen gas at 1 bar bubbling over it. Hydrogen gas at 1 bar is passed over the platinum. The platinum adsorbs the gas and catalyses the half reaction 2H+ plus 2 electrons giving H2. By convention its electrode potential is taken as zero at every temperature, because only potential differences between electrodes can be measured, not absolute potentials. Every standard electrode potential in the electrochemical series is therefore measured by making a cell with this electrode and finding the cell potential.

Hinglish: SHE mein platinum plate platinum black se coated hoti hai, 1 M acid mein doobi hoti hai, aur 1 bar H2 gas pass hoti hai. Iska half reaction 2H+ + 2e- → H2 hai. Sirf potential difference measure ho sakta hai, isliye SHE ka potential convention se zero rakha jaata hai; baaki sab potentials isi ke against nikalte hain.

Q4 Distinguish between conductivity and molar conductivity. Why does molar conductivity increase on dilution?

Answer (English): Conductivity is the conductance of a unit volume of the solution and is measured as the inverse of resistivity, in S cm^-1. It counts how well the solution as a whole carries current and therefore falls as the solution is diluted, because fewer ions remain in each unit volume. Molar conductivity is the conductivity divided by concentration and then scaled, giving S cm^2 mol^-1; it tells how well one mole of electrolyte conducts. On dilution the ions become separated by more solvent, interionic attractions fall, and a weak electrolyte dissociates further, so the molar conductivity rises. For strong electrolytes the rise is small and linear with the square root of concentration.

Hinglish: Conductivity ek unit volume ki conductance hai, unit S cm^-1; dilution par ions per volume kam hote hain isliye girti hai. Molar conductivity conductivity ko concentration se divide karke milti hai, unit S cm^2 mol^-1. Dilution par ions alag ho jaate hain aur weak electrolyte zyada dissociate hota hai, isliye molar conductivity badhti hai.

Q5 State Kohlrausch's law. How is it used to find the limiting molar conductivity of a weak electrolyte?

Answer (English): Kohlrausch's law states that at infinite dilution each ion migrates independently and contributes a definite amount to the molar conductivity, so the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions, each multiplied by its stoichiometric number. For a weak electrolyte such as acetic acid, direct extrapolation of the graph to zero concentration is unreliable. Instead, the limiting values of three strong electrolytes are combined algebraically, for example sodium acetate plus hydrochloric acid minus sodium chloride gives acetic acid. The ion that cancels is removed and the required sum remains. The same law also gives the limiting value of a sparingly soluble salt.

Hinglish: Kohlrausch law kehta hai infinite dilution par har ion alag contribute karta hai, isliye Λm° ions ke limiting values ka sum hoti hai. Weak electrolyte ka Λm° directly graph se nahi milta, isliye strong electrolytes ki values ko add-subtract karke nikaalte hain, jaise CH3COONa + HCl - NaCl = CH3COOH.

Q6 A 0.2 M KCl solution has conductivity 2.48 times 10^-2 S cm^-1. Calculate its molar conductivity and its degree of dissociation, given that the limiting ionic conductivities of K+ and Cl- are 73.5 and 76.5 S cm^2 mol^-1.

Answer (English): Molar conductivity is 1000 times the conductivity divided by the molarity, which is 1000 times 2.48 times 10^-2 divided by 0.2, giving 124 S cm^2 mol^-1. The limiting molar conductivity is the sum of the two ionic values, 73.5 plus 76.5, which is 150 S cm^2 mol^-1. The degree of dissociation is 124 divided by 150, which is 0.827. Potassium chloride is a strong electrolyte, so this value close to one is expected; the small gap is caused by interionic attraction at a finite concentration.

Hinglish: Molar conductivity = 1000 × 2.48×10^-2 / 0.2 = 124 S cm^2 mol^-1. Λm° = 73.5 + 76.5 = 150. Degree of dissociation α = 124/150 = 0.827, matlab strong electrolyte lagbhag poora dissociate hai aur bacha hua difference interionic attraction ki wajah se hai.

Q7 State Faraday's first law and calculate the mass of nickel deposited when a current of 5 A flows through Ni(NO3)2 solution for 20 minutes. Molar mass of Ni is 58.7 g mol^-1.

Answer (English): Faraday's first law states that the mass of substance liberated at an electrode is directly proportional to the quantity of electricity passed. The charge is 5 A times 1200 s, which is 6000 C. Nickel is deposited from Ni2+, so two electrons are needed per atom and z is 2. The amount deposited is 6000 divided by 2 times 96500, which is 0.0311 mol. Multiplying by the molar mass gives 0.0311 times 58.7, which is 1.83 g of nickel.

Hinglish: Faraday ka first law kehta hai deposited mass charge ke proportional hota hai. q = 5 × 1200 = 6000 C. Ni2+ ke liye z = 2, isliye moles = 6000/(2×96500) = 0.0311. Mass = 0.0311 × 58.7 = 1.83 g.

Q8 Why are the products of electrolysis different when aqueous silver nitrate is electrolysed with silver electrodes and with platinum electrodes?

Answer (English): With silver electrodes, the silver anode can itself be oxidised, and its oxidation potential is lower than that needed to oxidise water, so silver dissolves into the solution as Ag+. Silver ions are reduced at the cathode, so silver simply transfers from anode to cathode. With inert platinum electrodes, the anode cannot dissolve, so water is oxidised and oxygen gas is released while the cathode still deposits silver. The difference is a competition between the electrode material and water for oxidation at the anode.

Hinglish: Silver electrodes ke saath anode khud oxidise ho jaata hai aur Ag+ banta hai, cathode par silver deposit hota hai. Platinum inert electrode ke saath anode par water oxidise hota hai aur oxygen nikalti hai. Farq isliye hai kyunki anode par electrode material aur water ke oxidation ka competition hai.

Q9 Write the electrode reactions of a lead storage battery during discharge and during charging.

Answer (English): During discharge, lead at the anode is oxidised to lead sulphate while releasing two electrons, and lead dioxide at the cathode is reduced to lead sulphate in acid solution. The overall discharge reaction produces lead sulphate at both electrodes and water. During charging the external supply reverses both reactions: lead sulphate at the anode is oxidised back to lead dioxide and lead sulphate at the cathode is reduced back to lead. The sulphuric acid consumed in discharge is regenerated in charging, which is why the electrolyte density changes with the state of charge.

Hinglish: Discharge mein anode par Pb → PbSO4 (oxidation) aur cathode par PbO2 → PbSO4 (reduction) hota hai. Charging mein reaction reverse ho jaata hai: anode par PbSO4 wapas PbO2 banta hai aur cathode par PbSO4 wapas Pb. Sulphuric acid discharge mein use hota hai aur charging mein regenerate hota hai.

Q10 Explain, with one reaction for each, how zinc protects iron from rusting when iron is galvanised and how a sacrificial electrode works.

Answer (English): Rusting needs both oxygen and water and occurs through an electrochemical cell on the iron surface. In galvanising, a zinc layer coats the iron and acts as a physical barrier that keeps air and water away. Zinc also has a more negative standard potential than iron, so even if the coating is scratched, zinc is oxidised in preference to iron and protects the metal as a sacrificial anode. The same idea is used with blocks of magnesium or zinc fixed to iron pipelines and ship hulls; the blocks corrode while the iron stays intact, and they are replaced from time to time.

Hinglish: Galvanising mein zinc ki layer iron ko air aur paani se alag rakhti hai. Zinc ka potential iron se zyada negative hota hai, isliye coating tut jaane par bhi zinc pehle oxidise hota hai aur iron ko bachata hai. Isi principle par pipelines aur ships par magnesium ya zinc ke sacrificial electrode lagaye jaate hain jo khud corrode ho jaate hain.

The chapter quiz has five questions on cell notation, the Nernst equation, conductivity and the Faraday relations. Take the quiz after working the ten answers above, because each explanation points back to the step that matters in the numerical.