coloumb charge liberated 1 gm silver (Ag). If the charge is doubled then the amount of liberated Ag will be:
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 |
The cell reaction of an electrochemical cell is \(Cu^{2+}(C_{1}) + Zn \to Cu + Zn^{2+}(C_{2})\).
The change in free energy will be the function of:
\(1. \ ln (C_{1}+C_{2})\)
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. 139.5
2. 101.5
3. 203
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