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The quiz is designed for Electrochemistry chapter for JEE Main level.
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Question 1 of 15
1. Question
4 pointsGiven : Fe+2 + 2e- ⇌ Fe E0 = – 0.44 V
Co+2 + 2e-⇌ Co E0 = – 0.28 V
Ca+2 + 2e- ⇌ Ca E0 = – 2.87 V
Cu+2 + 2e- ⇌ Cu E0 = + 0.337 V
Which is the most electropositive metal?
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Question 2 of 15
2. Question
4 pointsA galvanic cell is made up to SHE as cathode and another hydrogen electrode as anode. The other hydrogen electrode should be immersed in which of the following solutions to get maximum value of emf?
(a) 0.1 M HCl (b) 0.1 M CH3COOH (Ka =10-5)
(c) 0.1 M H2SO4 (d) 0.1 M H3PO4 (Ka1 = 10-3)
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Question 3 of 15
3. Question
4 pointsWhich of the following is/are functions(s) of salt bridge
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Question 4 of 15
4. Question
4 pointsA cell contains two hydrogen electrodes. The negative electrode is in contact of 10-8 M hydrogen ions. The EMF of the cell is 0.236 V at 298 K. The concentration of hydrogen ions at the positive electrode is:
(a) 10-6 M (b) 10-4 M (c) 10-5 M (d) 10-3 M
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Question 5 of 15
5. Question
4 pointsFor the electrode Pb (PbSO4(s), SrSO4(s) ) SrCl2 (C molar), which of the following is correct representation of half cell reaction?
(a) Pb(s) + SrSO4(s) ⇌ PhSO4(s) + Sr2+ + 2e–
(b) Pb(s) + SrCl2 ⇌ PbCl2 + Sr2+ + 2e–
(c) Pb(s) + SrSO4 + PbSO4(s) ⇌ Pb2+ + Sr2+ + 2e–
(d) None of the above reaction is correct representation
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Question 6 of 15
6. Question
4 pointsFor the cell Ag(AgBr(s)) KBr (soln) ((Hg2Br2(s))Hg the measured e.m.f.s are 0.6839 v at 25oC and 0.07048 V at 30oC. What is the temperature coefficient of this cell?
(a) 4.2 x 10-4 v/oC (b) 1.3 x 10-5 v/oC
(c) 2.1 x 10-3 v/oC (d) 9.3 x 10-3 v/oC
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Question 7 of 15
7. Question
4 pointsThis cell reaction will be spontaneous when
(a) p1 = p2 = 1 atm (b) p1 > p2
(c) p1 < p2 (d)
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Question 8 of 15
8. Question
4 pointsCopper can be deposited from acidified CuSO4 and alkaline CuCN. If 5 amp. Current is passed for 5 min through both the solutions separately, which of the following is correct.
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Question 9 of 15
9. Question
4 pointsGiven electrode potentials are
Fe3+ + e → Fe2+ ; E0 = 0.771 V
I2 + 2e → 2I– ; E0 = 0.536 V
E0cell for the cell reaction; 2Fe3+ + 2I– → 2Fe2+ + I2; is
(a) (2 x 0.771 – 0.536) = 1.006 V (b) (0.771 – 0.5 x 0.536) = 0.503 V
(c) 0.771 – 0.536 = 0.235 V (d) 0.536 – 0.771 = -0.235 V
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Question 10 of 15
10. Question
4 pointsA hydrogen electrode placed in a buffer solution of CH3COONa and acetic acid in the ratios x:y and y:x has electrode potential values E1 volts and E2 volts respectively at 25oC. The pKa values of acetic acid is (E1 and E2 are oxidation potential)
(a) (b)
(c) (d)
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Question 11 of 15
11. Question
4 pointsIn an Electrolytic cell electrons flow from
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Question 12 of 15
12. Question
4 pointsConsider the following reactions:
Pb2+ (aq) + 2e– → Pb (s) E0 = -0.127 V
Cu+ (aq) + e– → Cu (s) E0 = +0.518 V
Which statement is true about a galvanic cell employing Pb, Cu, Pb2+ and Cu+?
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Question 13 of 15
13. Question
4 pointsTwo platinum electrodes are dipped in a solution of HCl, and H2 and Cl2 gases are made to bubble through them at constant pressure. One half cell is Pt, H2(g)|H3O+ (aq) with reaction ½ H2 + H2O → H3O+ + e–.
The electrode at which this half cell reaction would occur is
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Question 14 of 15
14. Question
4 pointsCu+ + e– → Cu, E0 = x1V, Cu+2 + 2e– → Cu, E0 = x2V
For Cu+2 + e– → Cu+, E0 will be
(a) x1 – 2x2 (b) x1 + 2x2
(c) x1 – x2 (d) 2x2 – x1
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Question 15 of 15
15. Question
4 pointsGiven Eo for Cu2+ → Cu+ is + 0.15 V and Cu+ → Cu is +0.50 V
Calculate Eo for Cu+2 → Cu
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