Electrochemistry · Question 55

The equilibrium constant of the reaction: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s); E = 0.46 V at 298 K is

1. 2.0 1010
2. 4.0 1010
3. 4.0 1015
4. 2.4 1010

The efficiency of a fuel cell is given by

1. G/ S
2. G/ H
3. S/ G
4. H/ G

A hypothetical electrochemical cell is shown below. A | A+(xM) || B+(yM) | B The emf measured is + 0.20 V. The cell reaction is

1. A + B+ A+ + B
2. A+ + B A + B+
3. A+ + e- A; B+ + e- B
4. the cell reaction cannot be predicted.

E Fe2+/Fe = -0.441 V and E Fe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe3+ 3Fe2+ will be

1. 0.111 V
2. 0.330 V
3. 1.653 V
4. 1.212 V

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge will be

1. 44.8 L
2. 22.4 L
3. 11.2 L
4. 5.6 L

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25 C. The equilibrium constant of the reaction would be

1. 2.0 1011
2. 4.0 1012
3. 1.0 102
4. 1.0 1010

The e.m.f. of a Daniell cell at 298 K is E1. [figure] When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2?

1. E1 > E2
2. E1 < E2
3. E1 = E2
4. E2 = 0 E1

On the basis of the information available from the reaction, 4/3 Al + O2 2/3 Al2O3, G = -827 kJ mol-1 of O2, the minimum e.m.f. required to carry out an 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

In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at 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. concentration of H+ ions is larger when Pt electrode is taken

Standard electrode potentials are Fe2+/Fe [ E = -0.44 V] and Fe3+/Fe2+ [ E = 0.77 V]; If Fe2+, Fe3+ and Fe blocks are kept together, then

1. Fe3+ increases
2. Fe3+ decreases
3. Fe2+/Fe3+ remains unchanged
4. Fe2+ decreases.