The incorrect statement about the solution of electrolytes is:

1.	Conductivity of solution depends upon the size of ions.
2.	Conductivity depends upon the viscosity of solution.
3.	Conductivity does not depend upon the solvation of ions present in solution.
4.	Conductivity of solution increases with temperature.

Using the data given below find out the strongest reducing agent.

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>;<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{0}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>;\hspace{0.17em}{E}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

1.  Cl^-

2.  Cr

3.  Cr³⁺

4.  Mn²⁺

Using the data given below find out the strongest oxidizing agent.

\(\mathrm{E_{Cr_{2} O_{7}^{2 -} / Cr^{3 +}}^{\ominus} = 1 . 33 V     ;     E_{Cl_{2} / Cl^{-}}^{\ominus} = 1 . 36 V \\ E_{Mn O_{4}^{-} / Mn^{2 +}}^{\ominus} = 1 . 51 V         ;    E_{Cr^{3 +} / Cr}^{\ominus} = - 0 . 74 V}\)

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>;<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{\mathrm{O}}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>;\hspace{0.17em}{E}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

Based on the given data, identify the option that correctly lists the order of reducing power.

1.  Cr³⁺ < Cl^– < Mn²⁺ < Cr

2.  Mn²⁺ < Cl^– < Cr³⁺ < Cr

3.  Cr³⁺ < Cl^– < Cr₂O₇^2– < MnO₄^–

4.  Mn²⁺ < Cr³⁺ < Cl^– < Cr

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V};<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{\mathrm{O}}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V};\hspace{0.17em}{E}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

Use the data given above to find out the most stable ion in its reduced form.

The most stable oxidized species among the following is: 
\(E_{{\mathrm{Cr}_2 \mathrm{O}_7^2}/ \mathrm{Cr}^{3+}}^{o} =1.33 \mathrm{~V} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{o}=1.36 \mathrm{~V} \)
\( E_{\mathrm{MnO_{4}}^{-} / \mathrm{Mn}^{2+}}^{o}=1.51 \mathrm{~V} ; E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=-0.74 \mathrm{~V}\)

The quantity of charge required to obtain one mole of aluminium from Al₂O₃ is :

1.  1 F

2.  6 F

3.  3 F

4.  2 F

The cell constant of a conductivity cell-

\(\Lambda _{m(NH_{4}OH)}^{o}\) is equal to -
1. \(\Lambda _{m(NH_{4}OH)}^{o} \ + \ \Lambda _{m(NH_{4}Cl)}^{o} \ - \ \Lambda _{m(HCl)}^{o}\)
2. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaOH)}^{o} \ - \ \Lambda _{m(NaCl)}^{o}\)
3. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaCl)}^{o} \ - \ \Lambda _{m(NaOH)}^{o}\)
4. \(\ \Lambda _{m(NaOH)}^{o} \ + \ \Lambda _{m(NaCl)}^{o}\ - \ \Lambda _{m(NH_{4}Cl)}^{o}\)