SELECT * FROM question_mgmt as q WHERE id=1784 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=58 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
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)
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