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Conductivity always decreases with decrease in concentration (on dilution of the solution) both for weak as much as for strong electrolytes. it is because the number of ions per unit volume that carry the current is a solution decreases on dilution.

Because number of ions decreases per unit volume.

(b) equivalent weight

(b) NH₄CI + ZnCl₂

(c) Zinc amalgamated with mercury

(b) Lithium – ion battery

(b) Fe₂O₃ .xH₂O 

(a) All the above

(b) specific conductance

(a) 1.10V

Cell potential = E_cathode – E_anode

= 0.34 – (- 0.76) 

= 0.34 + 0.76 

= 1.10V

(a) Fe is oxidised to Fe²⁺ and dissolved oxygen in water is reduced to OH^-2

The reciprocal of the resistance \(\Big(\frac{1}{2}\Big)\) is known as the conductance of an electrolytic solution. The SI unit of conductance is Ohm^-1 (or) Siemen (S)

The specific conductance is defined as the conductance of a cube of an electrolytic solution of unit dimensions. The SI unit of specific

k = C.\(\frac{l}{A}\)

conductance is Sm^-1

(d) Ohm^-1 m^-1

0.005M KCl > 0.01M KCI > 0.1M KCI > 0.25KCl > 0.5 KCI. 

Specific conductance and concentration of the electrolyte. So if concentration decreases, specific conductance increases.

1. AC current to prevent electrolysis of the solution

2. If we apply DC current to the cell the positive ions will be attracted to the negative plate and the negative ions to the positive plate. This will cause the composition of the electrolyte to change while measuring the equivalent conductance.

3. So DC current through the conductivity cell will lead to the electrolysis of the solution taken in the cell. To avoid such a electrolysis, we have to use AC current for measuring equivalent conductance.

(d) \(\frac{Equivalent\,mass}{96500}\)

(a) Nernst equation

(a) ΔG°° = -RTInk_eq

(b) Nernst equation

SHE Value is zero 

E_cell° = E_R° – E_L° 

= 0.34 – 0 = 0.34V 

The reduction potential of copper electrode = 0.34V

The standard electrode potential of SHE at 298K is 0.0 V

(b) Ca, K, Li

(a) – ve, > 1, + ve.

(d) Zinc has higher negative electrode potential than iron

E⁰_zn+|zn = – 0.76V and E⁰_Fe^2+|Fe = 0.44V. 

Zinc has higher negative electrode potential than iron, iron cannot be coated on zinc.

Zinc is more reactive than copper. Hence, it displaces copper from copper sulphate solution as follows 

Zn(s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s) 

So, we cannot store copper sulphate solution in a zinc pot.

(a) Ohm metre

(a) specific resistance

The unit of resistivity is Ω m

V = IR represents Ohm’s law

m = ZIt is represents Faraday’s first law.

1. Specific resistance (or) Resistivity 

2. Specific conductance, Kappa(k) 

3. Sm^-1

4.  Λ = \(\frac{k\times10^{-3}}{N}\)

5. decrease 

6. Kohlraush’s law 

7. Galvanic (or) Voltaic cell 

8. Electrolytic cell 

9. KCl (or) Na₂SO₄ 

10. Volt (V)

(b) charge carried by one mole of electrons

IF = 96500 C = 1 charge of mole of e^- = charge of 6.022 x 10²³ e^-

(a) 5F

7MnO₄^- + 5e^- → Mn²⁺ + 4H₂O

5 moles of electrons i.e., 5F charge is required.

(a) 5F

7MnO₄^- + 5e^- → Mn²⁺ + 4H₂O 

5 moles of electrons i.e., 5F charge is required.

A galvanic cell works basically in reverse to an electrolytic cell. The anode is the electrode where oxidation takes place, in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode.

The anode is also the electrode where metal atoms give up their electrons to the metal and go into solution. The electron left behind on it render ¡t effectively negative and the electron flow goes from it through the wire to the cathode.

Positive aqueous ions in the solution are reduced by the incoming electrons on the cathode. This why the cathode is a positive electrode, because positive ions are reduced to metal atoms there.

(b) Li – ion battery

(b) non – rechargeable

(b) (iii) only

(c) Aluminium

Au has the highest electrode potential

(a) Bakelite V

(a) Paste of kOH and ZnO

(c) Redox reactions

(b) Chemical energy

(b) Fuel cell

(b)  ii only

87.135 r cm, 0.01148 S cm^-1 , 229.6 S cm² mol^-1.

In lead storage cell, the oxidant (PbO₂), reduction (Pb) and their redox product (PbSO₄) are solids. Thus half cells need not to be in separate vessels; also they have common electrolyte solution of 40% of H₂SO₄. 

(a) Both A and R are correct and R is the correct explanation of A