i) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

(i) The half cell reactions are :
\(2Br^{-} \rightarrow Br_{2}+2e^{-}\)     \(E^{0}_{ox}=-1.09V\) ...(1)
\(2Fe^{3+}+2e^{-}\rightarrow2Fe^{2+}\)  \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}=+0.771V\) ..(2)
(ii) Adding (1) of (2) :
\(2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}\)   \(E^{0}_{cell}=?\) ...(3)
\(E^{0}_{cell}=E^{0}_{ox}+E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0}\)
= (-1.09 + 0.771)V
= -0.319V
(iii) E⁰_cell is – ve; \(\Delta G\) is +ve and the cell reaction is non spontaneous.
(iv) Hence Fe³⁺ cannot oxidises Br^- to Br₂.

Given : I = 1.608A;
t = 50 min = 50 \(\times\) 60 = 3000sec ; S = 250 mL;
C = 0.5 M ; \(\eta=100%\)%
i) Q = It     (I = Q / t)
= 1.608 x 3000 = 4824 Columb
No. of Faradays of electricity \(=\frac{4824}{96500}=0.04 \mathrm{~F}\)
ii) Electrolysis of CuSO₄
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}\)
\(\therefore\) 2 F electricity will deposit 1 mole of Cu²⁺ to Cu ​​​​
\(\therefore\) 0.05 F electricity will deposit 
\(x=\frac{1 \times 0.05}{2}=0.025\) mole
Initial No. of moles of Cu²⁺ in 250 ml of solution \(=\frac{0.5}{1000} \times 250 \) mole = 0.125 moles
\(\therefore\) No. of moles of Cu²⁺ after elettrolysis = 0.125 - 0.025 = 0.1 mole
\(\therefore\) concentration of Cu²⁺
\(=\frac{0.1}{250} \times 1000=0.4 \mathrm{M}\)

\(\kappa=\mathrm{C}\left(\frac{l}{\mathrm{~A}}\right)\) (or) k = 1/R.l/A
Where, C - Conductance
\(\frac{l}{A} \Rightarrow\) Cell constant (or)
Sp. Conductance \(\propto\) Cell constant
Sp. Conductance \(\propto\) Concentration
1. Sp. Conductance = concentration x cell constant
i) \(\kappa=0.1 \times 0.5=0.05=5 \times 10^{-2} \mathrm{Scm}^{-1}\)
ii) \(\kappa=0.1 \times 0.25=0.025=2.5 \times 10^{-2} \mathrm{Scm}^{-1}\)
2. So the first case will have greater specific conductance with cell constant 0.05.

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

1. \(\kappa=\mathrm{C}\left(\frac{1}{\mathrm{~A}}\right)\)
2. \(\operatorname{let} \frac{l}{A}=x\)
i) \(0.01 \mathrm{M} \mathrm{KCl}: \kappa=0.01 x=10^{-2} x\)
ii) \(0.005 \mathrm{M} \mathrm{KCl}: \kappa=0.005 x=5 \times 10^{-3} x\)
iii) \(0.1 \mathrm{M} \mathrm{KCl} \quad: \kappa=0.1 x=10^{-1} x\)
iv) \(0.25 \mathrm{M} \mathrm{KCl}: \kappa=0.25 x=2.5 \times 10^{-1} x\)
v) \(0.5 \mathrm{M} \mathrm{KCl}: \mathrm{K}=0.5 x=5 \times 10^{-1} x\)
\(\therefore 5 \times 10^{-1} x>2.5 \times 10^{-1} x>10^{-1} x>10^{-2} x>5 \times 10^{-3} x\)
\((ie) 0.5 \mathrm{M} \mathrm{KCl}>0.25 \mathrm{M} \mathrm{KCl}>0.1 \mathrm{M} \mathrm{KCl}>0.01 \mathrm{M} \mathrm{KCl}>0.005 \mathrm{M} \mathrm{KCl}\)

Electrochemical reaction at cathode is Ag⁺ + e^-  → Ag (reduction)
m = ZIT
Z = \(\frac{\text {molar mass of Ag}}{(96500)}\) = \(\frac{108}{1 \times 96500} \)
I = 2A
t = 20 x 60S = 1200 S
It = 2A  x 1200S = 2400C
m = \(\frac{108 gmol^{-1}}{96500 C mol^{-1}} \times\) 2400C
m = 2.68g

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

Metals having higher oxidation potential will liberate H₂ from H₂SO₄. Hence, the metal M₁ having +xV, oxidation potential will liberate H₂ from H₂SO₄.

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  and 
\((E^{0}_{red})_{Cu^{2+}|Cu} = 0.34\)
These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.