Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ΔrGø¸ and equilibrium constant of the reactions.
(i) Eø Cr3+ / Cr = - 0.74 V
Eø Cd2+ / Cd = - 0.40 V
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
Eø = EøR - EøL
= -0.40 - (-0.74)
= + 0.34 V
ΔrGø = -nFEøcell
In the given equation,
n = 6
F = 96487 C mol - 1
Eøcell = +0.34 V
Then, ΔrGø = - 6 × 96487 C mol - 1 × 0.34 V
= - 196833.48 CV mol - 1
= - 196833.48 J mol - 1
= - 196.83 kJ mol - 1
Again,
ΔrGø = - RT ln K
= 34.496
K = antilog (34.496)
= 3.13 × 1034
(ii) Eø Fe3+ / Fe2+ = 0.77 V
Eø Ag+ / Ag = 0.80 V
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
Eø = EøR - EøL
= 0.80 - 0.77
= 0.03 L
Here, n = 1.
Then, ΔrGø = -nFEøcell
= - 1 × 96487 C mol - 1 × 0.03 V
= - 2894.61 J mol - 1
= - 2.89 kJ mol - 1
Again, ΔrGø = - 2.303 RT ln K
= 0.5073
K = antilog (0.5073)
= 3.2 (approximately)
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
I | 0.1 | 0.1 |
6.0 × 10 - 3 |
II | 0.3 | 0.2 |
7.2 × 10 - 2 |
III | 0.3 | 0.4 |
2.88 × 10 - 1 |
IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 4: Calculate the standard cell potentials of galvanic cells in which the following reactions take place....
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