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

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V} ; E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{0}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V} ; E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

1.  Cl^-

2.  Cr

3.  Cr³⁺

4.  Mn²⁺

 

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

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V} ; E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{0}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V} ; E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

1.  Cl^-

2.  Mn²⁺

3.  MnO₄^-

4.  Cr³⁺

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V} ; E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{0}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V} ; E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

Using the data given above find out in which option the order of reducing power is correct.

1.  Cr³⁺ < Cl^- < Mn²⁺ < Cr

2.  Mn²⁺ < Cl^- < Cr³⁺ < Cr

3.  Cr³⁺ < Cl^- < Cr₂O₇^2- < MnO₄^-

4.  Mn²⁺ < Cr³⁺ < Cl^- < Cr

$\begin{array}{l}{E}_{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}/{\mathrm{Cr}}^{3+}}^{⊝}=1.33\mathrm{V} ; E_{{\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}}^{\ominus }=1.36\mathrm{V}\\ E_{\mathrm{Mn}{\mathrm{O}}_4^{-}/{\mathrm{Mn}}^{2+}}^{⊝}=1.51\mathrm{V} ; E_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\ominus }=-0.74\mathrm{V}\end{array}$

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-

The correct statement about charging of the lead storage battery is:

1. PbSO₄ anode is reduced to Pb.

2. PbSO₄ cathode is reduced to Pb.

3. PbSO₄ cathode is oxidised to Pb.

4. PbSO₄ anode is oxidised to PbO₂.

\(\Lambda _{m(NH_{4}OH)}^{o}\) 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. \(\ \Lambda _{m(NaOH)}^{o} \ + \ \Lambda _{m(NaCl)}^{o}\ - \ \Lambda _{m(NH_{4}Cl)}^{o}\)

The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution  is represented by : 

1. Na⁺(aq) + e^- ⟶ Na(s) ; \(E_{cell}^{o} \ = \ -2.71 \ V \)

2. 2H₂O(l) ⟶ O₂(g) + 4H⁺(aq) + 4e^- ; \(E_{cell}^{o} \) = 1.23 V

3. H⁺(aq) + e^- ⟶ \(\frac{1}{2}\)H₂(g) ; \(E_{cell}^{o} \) = 0.00 V

4. Cl^-(aq) ⟶ \(\frac{1}{2}\)Cl₂(g) + e- ; \(E_{cell}^{o}\) $= 1.\mathrm{36 }\mathrm{V}$