ElectrochemistryHard
Question
Given : Hg22+ + 2e → 2Hg , Eo = 0.789 V & Hg2+ + 2e → Hg, Eo = 0.854 V, calculate the equilibrium constant for Hg22+ → Hg + Hg2+.
Options
A.3.13 × 10-3
B.3.13 × 10-4
C.6.26 × 10-3
D.6.26 × 10-4
Solution
Hg22+ + 2e- → 2Hg , 0.789 Volt
Hg → Hg2+ + 2e-, -0.854 Volt
Hg22+ → Hg + Hg2+, -0.065 Volt
ᐃG = - 2 × (- 0.065) × 96500 = -8.314 × 298 ln Keq. ;
Keq.= 6.3 × 10-3
Hg → Hg2+ + 2e-, -0.854 Volt
Hg22+ → Hg + Hg2+, -0.065 Volt
ᐃG = - 2 × (- 0.065) × 96500 = -8.314 × 298 ln Keq. ;
Keq.= 6.3 × 10-3
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