The value of E⁰ cell for the following reaction is:
\(Cu^{2+}+ Sn^{2+}\to Cu +Sn^{4+ } \)

(Given, equilibrium constant is 10⁶)

For the disproportionation of copper:

2Cu⁺ → C u²⁺ + C u$,$ $E°$ is:
(Given $E°$ 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

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:

On the basis of the information available from the reaction:

$\frac{4}{3}\mathrm{Al}+{\mathrm{O}}_2\to \frac{2}{3}{\mathrm{Al}}_2{\mathrm{O}}_3,$ $∆\mathrm{G}=-827$ $\mathrm{KJ}$ ${\mathrm{mol}}^{-1}$

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

The EMF of a Daniel cell at 298 K is E₁  Zn|ZnSO₄(0.01 M) || CuSO₄(1.0 M)|Cu.
When the concentration of ZnSO₄ is 1.0 M and that of CuSO₄ is 0.01 M, the EMF is changed to E₂. The correct relationship between E₁ and E₂ is: