The correct statement among the following options is: 

1.	Ecell and ∆rG of cell reaction are both extensive properties.
2.	Ecell and ∆rG of cell reaction are both intensive properties.
3.	Ecell is an intensive property while ∆rG of cell reaction is an extensive property.
4.	Ecell is an extensive property while ∆rG of cell reaction is an intensive property.

The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called : 

1. Cell potential.

2. Cell emf.

3. Potential difference.

4. Cell voltage.

The incorrect statement about an inert electrode in a cell is:

An electrochemical cell can behave like an electrolytic cell when -

1. E_cell = 0

2. E_cell > E_ext

3. E_ext > E_cell

4. E_cell = E_ext

The incorrect statement about the solution of electrolytes is:

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

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

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

1.  Cl^- 

2.  Cr³⁺ 

3.  Cr

4.  Mn²⁺ 

From the data given below, the most stable oxidized species is : 
\(\begin{aligned} &E_{{\mathrm{Cr}_2 \mathrm{O}_7^2}/ \mathrm{Cr}^{3+}}^{o} =1.33 \mathrm{~V} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{o}=1.36 \mathrm{~V} \\ &E_{\mathrm{MnO_{4}}^{-} / \mathrm{Mn}^{2+}}^{o}=1.51 \mathrm{~V} ; E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=-0.74 \mathrm{~V} \end{aligned}\)

1. Cr³⁺ 

2. MnO₄^-

3. Cr₂O₇^2-

4. Mn²⁺