The cell that will measure the standard electrode potential of a copper electrode is:

1.	Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, 1M) || Cu2+(aq, 1M) | Cu
2.	Pt(s) | H2(g, 1 bar) | H+(aq, 1M) || Cu2+ (aq, 2M) | Cu
3.	Pt(s) | H2(g, 1 bar) | H+(aq, 1M)|| Cu2+ (aq, 1M) | Cu
4.	Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, \(1 \over 10\)M) || Cu2+(aq, 1M) | Cu

The electrode potential for Mg electrode varies according to the equation

\(E_{Mg^{2+}/Mg}\ = \ E_{Mg^{2+}/Mg}^{o} \ - \ \frac{0.059}{2}log\frac{1}{[Mg^{2+}]}\)

The graph of E_{Mg²⁺ / Mg} vs log [Mg²⁺] among the following is:

1.		2.
3.		4.

The correct statement among the following options is: 

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