ElectrochemistryHard
Question
At 298K the standard free energy of formation of H2O(l) is -237.20 kJ/mole while ᐃGo of its ionisation into H+ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be
H2(g,1 bar) | H+ (1M) || OH- (1M) | O2 (g, 1bar)
H2(g,1 bar) | H+ (1M) || OH- (1M) | O2 (g, 1bar)
Options
A.0.40 V
B.0.81 V
C.1.23 V
D.-0.40 V
Solution
Cell reaction Cathode : H2O(l) + 
O2(g) + 2e- → 2OH-(aq.)
Anode : H2(g) → 2H+ (aq.) + 2e-
____________________________________
H2O(l) +
O2(g) + H2(g) → 2H+(aq.) + 2OH-(aq.)
Also we have
H2(g) +
O2(g) → H2O(l) ᐃGof = - 237.2 kJ/mole
H2O(l) → H+(aq.) + OH- (aq.) ᐃGo = 80 kJ/mole
Hence for cell reaction
ᐃGo = -77.20 kJ/mole So, Eo = -
= 0.40 V
O2(g) + 2e- → 2OH-(aq.)Anode : H2(g) → 2H+ (aq.) + 2e-
____________________________________
H2O(l) +
O2(g) + H2(g) → 2H+(aq.) + 2OH-(aq.)Also we have
H2(g) +
O2(g) → H2O(l) ᐃGof = - 237.2 kJ/moleH2O(l) → H+(aq.) + OH- (aq.) ᐃGo = 80 kJ/mole
Hence for cell reaction
ᐃGo = -77.20 kJ/mole So, Eo = -
= 0.40 VCreate a free account to view solution
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