What is the mass of silver deposited when the charge passed is doubled, if a charge of \(10^5\) coulomb deposits 1 g of Ag?
| 1. | 1 gm | 2. | 2 gm |
| 3. | 3 gm | 4. | 4 gm |
The concentration of solution will change when it is placed in a container which is made of:
| 1. | Al | 2. | Cu |
| 3. | Ag | 4. | None |
Which of the following quantities does the change in Gibbs free energy (ΔG) depend upon, for the electrochemical cell reaction: \(Cu^{2+}(C_{1}) + Zn \to Cu + Zn^{2+}(C_{2})\)
\(1. \ ln (C_{1}+C_{2})\)
2. \(ln (\frac{C_{2}}{C_{1}})\)
\(3. \ ln C_{2}\)
\(4. \ lnC_{1}\)
The value of E0 cell for the following reaction is:
\(Cu^{2+}+ Sn^{2+}\to Cu +Sn^{4+ } \)
(Given, equilibrium constant is 106)
| 1. | 0.17 | 2. | 0.01 |
| 3. | 0.05 | 4. | 1.77 |
For the disproportionation of copper:
2Cu+ → C u2+ + C u is:
(Given for Cu+2/Cu is 0.34 V & Eº for Cu+2/Cu+ is 0.15 V )
1. 0.49 V
2. – 0.19 V
3. 0.38 V
4. – 0.38 V
A cell reaction become spontaneous when:
1. ∆Gº is negative
2. ∆Gº is positive
3. is positive
4. is negative
At infinite dilution, equivalent conductances of Ba+2 & Cl– ions are 127 & 76 ohm–1cm–1 eq–1 respectively. Equivalent conductance (ohm–1cm–1 eq–1) of BaCl2 at infinite dilution is:
1. 203
2. 101.5
3. 139.5
4. 279
In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at the cathode while with Hg cathode it forms sodium amalgam because:
| 1. | Hg is more inert than Pt |
| 2. | More voltage is required to reduce H+ at Hg than at Pt |
| 3. | Na is dissolved in Hg while it does not dissolve in Pt |
| 4. | The concentration of H+ ions is larger when the Pt electrode is taken |
In the silver plating of copper, K[Ag(CN)2] is used instead of AgNO3. The reason is:
| 1. | A thin layer of Ag is formed on Cu |
| 2. | More voltage is required |
| 3. | Ag+ ions are completely removed from the solution |
| 4. | Less availability of Ag+ ions, as Cu can not displace Ag from [Ag(CN)2]– ion |
Consider the following reaction:
\(\frac{4}{3} \mathrm{Al}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3(\mathrm{~s})\)
The minimum e.m.f. required to carry out the electrolysis of Al2O3 is:
(F = 96500 C mol–1)
1. 2.14 V
2. 4.28 V
3. 6.42 V
4. 8.56 V