Questions - Answers
2 Marks questions:
1. The standard e.m.f. of Daniel Cell is 1.1 V, then calculate standard Gibbs energy for the Daniel Cell?
A: n = 2, F = 96500 C/mole, E°Cell = 1.1 V
ΔrG° = −nF E°Cell = −2 × 96500 × 1.1
= −21,230 Joules/mole
2. Determine the value of Kc for the following reaction.
Ni (s) + 2 Ag+ (aq.) → Ni+2 (aq.) + 2 Ag (s)
E° = 1.05 V
∴ Kc = 1035.59 = Anti log of 35.59 = 3.8 × 1035
3. Give the cell representation & calculate Ecell for the reaction.
Mg (s) + 2 Ag+ (0.0001 M) → Mg+2 (0.130 M) + 2 Ag (s), E° = 3.17 V
A: Mg |Mg+2 (0.130 M) || Ag+ (0.0001 M) | Ag
= 3.17 − 0.21
= 2.96 V
4. Calculate the potential of hydrogen electrode placed in a solution on of pH = 10.
A:
5. Asolution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of Copper deposited at the Cathode?
4 Marks questions:
1. What is electrolysis? Give the products obtained at the Pt anode and cathode respectively in the electrolysis of (a) fused NaCl (b) aqueous solution of K2SO4.
A: The process of decomposition of a chemical compound (either in molten state or in aqueous state) into its constituent elements under the influence of EMF is called electrolysis.
(a) at cathode: Na is produced; at anode: Cl2 is liberated.
(b) at cathode: H2 is liberated; at anode : O2 is liberated.
2. What is Standard Hydrogen Electrode (SHE)? Give the construction & working of SHE with a neat diagram.
A: The electrode formed by bubbling H2 gas at 1 atm pressure over the Platinum electrode (coated with Pt black) present in 1 M HCl solution is called SHE. E° for SHE is zero. IUPAC representation of SHE is
Pt (s) H2 (g) H+ (1 M)
By taking SHE as reference half cell (as Anode) and other half cell as Cathode (right side), we can get the reduction potential of the other half cell.
E°cell = E°R.H.S. − E°L.H.S.
= E°R.H.S. − 0
= E°R.H.S.
3. Calculate Kc for the reaction at 298 K.
Zn (s) + Cu+2 (aq.) ⇌ Zn+2 (aq.) + Cu (s)
(ΔGo = −212.3 K.J./ mole)
4. What are Galvanic cells? Explain the working of a galvanic cell with neat sketch taking Daniel cell as example.
A: Galvanic Cell: A device that converts chemical energy produced in a redox reaction into electrical energy is called Galvanic cell.
Anode: Zn rod dipped in ZnSO4 (aq.) solution acts as anode (taken on left hand side). Oxidation takes place at anode.
Zn (s) → Zn+2 (aq.) + 2 
(oxidation)
Cathode: Cu rod dipped in CuSO4 (aq.) solution acts as Cathode (taken on right hand side). Reduction takes place at Cathode.
Cu+2 (aq.) + 2 → Cu (s) (reduction)
Salt bridge: U shaped glass tube filled with KCl and Agar agar gel. This bridge connects both the electrolytic solutions. It helps in not accumulating charges, so that electricity is produced continuously. Zn rod & Cu rod are connected externally.
Cell reaction: Zn (s) + Cu+2 (aq.) → Zn+2 (aq.) + Cu (s)
Cell representation: Zn (s) | Zn+2 (aq.) || Cu+2 (aq.) | Cu (s)
e.m.f. of the cell can be calculated by using the formula.
5. State and explain Faraday's first law of electrolysis.
A: Faraday's first law: The mass of substance (m) deposited (or dissolved or liberated) at an electrode during electrolysis is directly proportional to the quantity of electricity (Q) passed through fused or aqueous electrolyte.
m ∝ Q
m ∝ C.t
m = C.e.t
where C = current (in amperes) t = time (in seconds)
Electro chemical equivalent (e): The mass of substance deposited (or dissolved or liberated) at an electrode when 1 coulomb of electricity is passed through an electrolyte.
Chemical equivalent (E): The mass of substance deposited (or dissolved or liberated) at an electrode when 96500 coulombs (called 1 Faraday) of electricity is passed through an electrolyte.
Faraday (F): The quantity of electric charge carried by one mole of electrons.
1 Faraday = 96500 coulombs
6. State Faraday's second law of electrolysis. Certain quantity of electricity liberates 0.504 grams of H2. Calculate the amount of Ag deposited for the passage of same quantity of electricity.
A: Faraday's Second Law: When the same quantity of electricity is passed through different cells containing fused or aqueous electrolytes, connected in series, the mass of substances deposited (or dissolved or liberated) at electrodes are in the ratio of their equivalent weights.
7. (a) Calculate the amount of Ag deposited, if 48250 coulombs of electricity is passed through AgNO3.
(b) Equivalent conductance of NaCl, HCl, C2H5, COONa at infinite dilution are 126.45, 426.16, 91 ohm-1 cm2 respectively. Calculate the equivalent conductance of C2H5COOH.
A: (a) W = e. Q = 
× 48250 = 54 grams of Ag
= 91 + 426.16 - 126.45
= 390.71 ohm-1 cm2
8. What is Kohlrausch law? Give its applications.
A: Kohlrausch's law: The equivalent conductance of an electrolyte at infinite dilution (Λ∞ ) is equal to the algebraic sum of equivalent conductance of cation (λ+∞ ) and anion (λ-∞).
Λ∞ Electrolyte = λ+∞ + λ-∞
Applications: 1) In the calculation of ''Λ∞'' of weak electrolyte.
2) To calculate "α" of weak electrolytes.
3) To calculate solubilities of sparingly soluble salts like BaSO4.
9. What is Nernst equation?
Calculate the electrode potential of Zn dipped in 0.04 M ZnSO4 solution (E° for Zn = − 0.76 V).
A: Nernst equation explain the effect of concentration of electrolyte on emf of the cell.
* Nernst equation for cation (metal) electrode E = E°+ 
log C
* Nernst equation for anion (non metal) electrode E = E° - log C
E = E° + log C
= - 0.76 + 
log (0.04)
= - 0.76 + × log (10-2 × 4) = - 0.801 volts.
Chemical Kinetics
8 Marks questions:
1. Define rate of chemical reaction. Explain the factors that influence the rate of a reaction.
A: The change (decrease) in molar concentration of reactants or the increase in molar concentration of products in Unit time is known as the rate of a reaction.
Units of reaction rate: Moles lit−1 sec−1.
Rate = 
The relation between the rates of the reactants P, Q and the products R, S for the reaction
x P + y Q → mR + nS is
Factors effecting reaction rates:
1. The chemical nature of the reactants: Rate of a reaction mainly depends upon the chemical nature of the reactants. A reaction taking place between ionic compounds is faster than the reaction taking place between covalent compounds. For example the reaction
between Ag+ (aq.) and Cl- (aq.) occurs very fast to give AgCl (white precipitate).
NaCl (aq.) + AgNO3 (aq.) → AgCl ↓ + NaNO3 (aq.)
The reactions taking place between covalent compounds involve breaking of existing bonds and making of new bonds. Hence such reactions take much time. Formation of NH3 from N2 and H2
N2 (g) + 3 H2 (g) 
2 NH3 (g)
2. Effect of temperature: According to Arrhenius equation, the rate of a reaction increases with increase in the temperature of the reaction. For every 10°C rise of temperature, specific rate or rate constant gets doubled.
Temperature coefficient = 
Arrhenius equation = 
where A = Arrhenius frequency factor
R = Gas constant
Ea = Activation energy
k1, k2 = Rate constants at = T1K, T2
According to collision theory of reaction rates, with the rise of temperature, the number of activated collisions increases to increase the rate of a reaction.
3. Effect of catalyst: Reactions take place slowly at room temperature. But in presence of a positive catalyst (the substance that increases the rate of a reaction without being consumed during the course of the reaction), the path of the reaction changes in which activation energy is low and the rate of the reaction is more.
e.g.: 2 KClO3 
2 KCl + 3 O2
2. Give a detailed account of the collision theory of reaction rates of bimolecular gaseous reactions.
A: * This theory was proposed by Arrhenius and developed by Max Trautz & William Lewis to explain the rates of gaseous reactions.
* It is based on kinetic theory of gases.
* According to this theory the reactant molecules are assumed to be hard spheres.
* Collisions must takes place between reactants (molecules) to occur a reaction.
* Collisions are possible if 2 or more molecules are present.
* The minimum energy must be possessed by the reactant molecules to give products is called threshold energy (ET).
* Molecules having other than threshold energy are called "normal molecules".
* Collisions between normal molecules are called "normal collisions", which do not lead the chemical reaction, hence products are not formed.
* In addition to the normal molecules, the molecules must aquire some extra amount of energy, called "activation energy" (Ea).
Ea = ET − ER
* The molecules possesing threshold energy are called activated molecules.
* Proper orientation of reactant molecules lead to bond formation and products are formed.
* Collisions occuring between activated molecules with proper orientation are called "effective collisions".
* Effective collisions are less in number, when compared to normal collisions.
* The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). Activation energy also affects the rate of a reaction.
For a bimolecular elementary reaction
A + B → Products
Rate = P.ZAB. e−Ea/RT
ZAB = Collision frequency of reactants A & B
P = Probability (Steric) factor
Ea = Activation energy
R = Universal gas constant
* Rate of a reaction is determined by Ea & proper orientation of the molecules.
e.g.: formation of CH3OH from CH3Br
Drawbacks:
* It considers atoms or molecules as hard spheres.
* It ignores structural aspect of molecules.
4 Marks Questions:
1. Distinguish between order and molecularity of a reaction.
A:
2. Give the relation between the rate constant (k) and half life (t1/2) for a first order reaction. The rate constant of a first order reaction is 0.693 sec-1. What is its t1/2?
A:
= 1 second.
3. Derive the integrated rate equation for the first order reaction.
A: Rate of the reaction = k [Reactants]1
R = reactants, P = products
R → P
Integrating both sides
ln R = −kt + I → (1)
I = Integration constant
When t = 0, R = [R]0
[R]0 = initial concentration of the reactant.
ln [R]0 = −k × O + I
∴ I = ln [R]0 → (2)
By Substituting (2) in (1)
ln R = −kt + ln [R]0
where a = initial concentration; x = concentration of the reactant consumed.
2 Marks Questions
1. Write the equation for the rate of the reaction.
5 Br− (aq.) + BrO3− (aq.) + 6 H+(aq.) → 3 Br2 (aq.) + 3 H2O (l)
2. Give 2 examples for zero order reactions.
3. Show that in the case of 1st order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion. (log 2 = 0.3010)
4. Sucrose decomposes in acid solution into glucose and fructose according to 1st order rate law, with 
hours. What fraction of sample of sucrose remains after 8 hours?
6. In a 1st order reaction, the concentration of the reactant is reduced from 0.6 mol/L to 0.2 mol/L in 5 min. Calculate the rate constant.
7. The rate constant of a reaction is doubled when the temperature is raised from 298 k to 308 k. Calculate the activation energy.
