For the hypothetical electrochemical cell given below:

A | A⁺(x M) || B⁺(y M) | B .

The measured cell emf is +0.20 V. Find the overall cell reaction.

1. A+ + B → A + B+

2.  A+ + e- → A ; B+ + e- → B

3. The cell reaction cannot be predicted.

4. A + B+ → A+ + B

Subtopic:  Electrochemical Series | Nernst Equation |
 77%
Level 2: 60%+
AIPMT - 2006
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The voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn2+ ions.

3. Increase in the concentration of Ag+ ions.

4. None of the above.

Subtopic:  Nernst Equation |
 63%
Level 2: 60%+
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The change in reduction potential of a hydrogen electrode when its solution initially
at pH = 0 is neutralised to pH = 7, is a/an:

1. Increase by 0.059 V 2. Decrease by 0.059 V
3. Increase by 0.41 V 4. Decrease by 0.41 V
Subtopic:  Electrode & Electrode Potential | Nernst Equation |
 55%
Level 3: 35%-60%
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By how much will the potential of half cell Cu2+| Cu change if the solution is diluted to 100 times at 298 K. It will -

1. Increase by 59 mV 2. Decrease by 59 mV
3. Increase by 29.5 mV 4. Decrease by 29.5 mV
Subtopic:  Nernst Equation |
 58%
Level 3: 35%-60%
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Calculate the EMF of the following concentration cell at 298 K:

Pt | H₂(1 atm) | H⁺(0.02 M) || H⁺(0.01 M) | H₂(1 atm) | Pt

1. - 0.017 V

2. 0.0295 V

3. 0.10 V

4. 0.059 V

Subtopic:  Nernst Equation |
 63%
Level 2: 60%+
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In the electrochemical cell:

Zn|ZnSO4(0.01 M) || CuSO4(1.0M),Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 is changed to 0.01 M, the emf changes to E2. The relationship between E1 and E2 is : 
( Given, \(\frac{R T}{F}\)= 0.059)

1. E1 = E2

2. E1 > E2

3. E1 < E2

4. E2 = 0 \(\neq\)E1

Subtopic:  Electrode & Electrode Potential | Nernst Equation |
Level 4: Below 35%
NEET - 2017
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The pressure of H2 required to make the potential of H- electrode zero in pure water at 298 K is:

1. 10–12  atm 2. 10–10  atm
3. 10–4  atm 4. 10–14 atm
Subtopic:  Nernst Equation |
 69%
Level 2: 60%+
NEET - 2016
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For the cell, Ti/Ti+(0.001M)||Cu2+(0.1M)|Cu, Ecello at

25 °C is 0.83 V. Ecell can be increased :

1. By increasing [Cu2+]

2. By increasing [Ti+]

3. By decreasing [Cu2+]

4. None of the above.

Subtopic:  Nernst Equation | Faraday’s Law of Electrolysis |
 73%
Level 2: 60%+
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Consider the given cell:

Pt(s) | H₂(g, 1 bar) | H⁺ (0.030 M) || Br⁻ (0.010 M) | Br₂(l) | Pt(s)

If the concentration of Br⁻ ions is doubled and the concentration of H⁺ ions is reduced to half of its initial value,
how will the emf of the cell change?

1. Two times
2. Four times
3. Eight times
4. Remains the same

Subtopic:  Nernst Equation |
 56%
Level 3: 35%-60%
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The electrode potential of Cu electrode dipped in 0.025 M CuSO4 solution at 298 K is:

(standard reduction potential of Cu = 0.34 V)

1. 0.047 V

2. 0.293 V

3. 0.35 V

4. 0.387 V

Subtopic:  Nernst Equation |
 63%
Level 2: 60%+
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