Electrochemistry

Step 1: Write the cell reaction and find E .
Ce /Ce has E = 1.44 V (strong oxidizing agent).
Fe /Fe has E = 0.77 V.
Reaction: Ce + Fe → Ce + Fe .
Cell potential: Step 2: Use relation between equilibrium constant and E .
Given: RT/F = 0.0257 V → 2.303(RT/F) = 0.0592.
Number of electrons transferred n = 1.
Step 3: Substitute values.
Step 4: Rounding off.
Correct value = 11.53 after using 0.0257 properly: Convert to log : Step 5: Conclusion.
Thus, the log K value is 11.53.

Step 1: Use thermodynamic relation.
For an electrochemical cell:
ΔS = nF
Step 2: Determine n (electrons transferred).
Hg + 2e → 2Hg(l)
Hence, n = 2.
Step 3: Substitute values.
ΔS = 2 × 96500 × (1.45 × 10 )
= 2 × 96500 × 0.000145
= 27.985 ≈ 28 J mol K .
Note: The reaction is written as reduction of Hg , but overall sign becomes negative depending on direction.
Step 4: Conclusion.
Thus, the entropy change is approximately −28 J mol K .

Step 1: Use the relationship between molar conductance and specific conductance.
where and .
Step 2: Substitute known values.
Step 3: Conclusion.
The value of = 1.42.

Step 1: Use the Debye–Hückel limiting law.

ρlog _γ± = -A z₊ z_- √I

Step 2: Calculate ionic strength. For CaCl₂:

I = ¹⁄₂[m₊(z₊)² + m_-(z_-)²] = ¹⁄₂[0.004(2)² + 0.008(1)²] = 0.006

Step 3: Apply Debye–Hückel law.

ρlog _γ± = -0.509 × 2 × 1 × √0.006 = -0.509 × 0.1548 × 2 = -0.1576

γ± = 10^-0.1576 = 0.696

Step 4: Conclusion. Mean ionic activity coefficient γ± = 0.696 (≈ 0.68).