Q. No.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 EMg2+ / Mg vs log [Mg2+] among the following is:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
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:
| 1. | It does not participate in the cell reaction. |
| 2. | It provides a surface either for oxidation or for the reduction reaction. |
| 3. | It provides a surface for the conduction of electrons. |
| 4. | It provides a surface for redox reaction. |
An electrochemical cell can behave like an electrolytic cell when -
1. Ecell = 0
2. Ecell > Eext
3. Eext > Ecell
4. Ecell = Eext
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 oxidizing agent.
\(\mathrm{E_{Cr_{2} O_{7}^{2 -} / Cr^{3 +}}^{\ominus} = 1 . 33 V ; E_{Cl_{2} / Cl^{-}}^{\ominus} = 1 . 36 V \\ E_{Mn O_{4}^{-} / Mn^{2 +}}^{\ominus} = 1 . 51 V ; E_{Cr^{3 +} / Cr}^{\ominus} = - 0 . 74 V}\)
1. Cl–
2. Mn2+
3. MnO4–
4. Cr3+
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– < Cr2O72– < MnO4–
4. Mn2+ < Cr3+ < Cl– < Cr
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