(a) Calculate ∆G° and log Kc for the following reaction at 298 K:
2Cr(s) + 3Fe2+(aq) → 2Cr3+(aq) + 3Fe(s)
Given: E°cell = 0.30 V
(b) Using the E° values of A and B, predict which is better for coating the surface of iron [E° (Fe2+|Fe) = –0.44 V] to prevent corrosion and why?
Given: E°(A2+|A) = –2.37 V : E°(B2+|B) = –0.14 V
OR
(a) The conductivity of 0.001 mol/L solution of CH3COOH is 3.905 × 10-5 S/cm. Calculate its molar conductivity and degree of dissociation.
Given: λ° (H+) = 349.6 S cm2 mol-1 and λ° (CH3COO-) = 40.9 S cm2 mol-1.
(b) What type of battery is lead storage battery? Write the overall reaction occurring in lead storage battery.
(a) Given:
2Cr(s) + 3Fe2+(aq) → 2Cr3+(aq) + 3Fe(s)
E°cell = 0.30 V
T = 298 K

And n (no. of electrons) = 6
Substituting the values in ΔG° = -nFE°cell
ΔG° = - 6
96500
0.30
= -173700 J/mol
= -173.7 KJ/mol
Now, log Kc = ![]()
log Kc = ![]()
= 30.50
(b) We have to coat the surface of iron having E° (Fe2+|Fe) = –0.44 V to prevent corrosion. Corrosion is the phenomenon of oxidation of iron, if we will coat the surface of iron with a substance having higher oxidation potential than that of iron, then that substance will be oxidized instead of iron. Thus, substance A will be coated on iron.
Oxidation potential of Fe: 0.44
Oxidation potential of A: 2.37
Oxidation potential of B: 0.14
OR
(a) Given, K, conductivity of CH3COOH = 3.905
10-5 Scm-1,
Molarity of CH3COOH, M = 0.001 M,
(H+) = 349.6 Scm2mol-1 and
⋀°(HCOO-) =40.9 Scm2mol-1
Molar conductance, ⋀m =
Scm2mol-1
⋀m =
= 39.05 Scm2mol-1
According to Kohlrauch’s Law, limiting molar conductivity of an electrolyte can be represented as sum of individual contribution of cation and anion of electrolyte.
Therefore, limiting molar conductivity of HCOOH can be written as,
⋀°m (CH3COOH) = ⋀°m (CH3COO-) + ⋀°m (H+)
= 40.9 Scm2mol-1 + 349.6 Scm2mol-1
= 390.5 Scm2mol-1
Degree of dissociation,![]()
= 0.1
Molar conductivity and degree of dissociation of ethanoic acid is 39.05 Scm2mol-1 and 0.1 respectively.
(b) Lead storage battery is a secondary battery. The overall reaction occurring in lead storage battery is
Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O
Couldn't generate an explanation.
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