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  • At 298K the standard free energy of formation of H2O(l) is –237.20 kJ/mole while that of its ionisation into H+ ion and hydroxyl ions is 80 kJ/mole, then the emf of the

At 298K the standard free energy of formation of H2O(l) is –237.20 kJ/mole while that 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 (in V).

(Take F = 96500 C)

H2 (g, 1 bar) | H+ (1M)| |OH- (1M) | O2 (g, 1 bar)

(Response should be upto only two decimal digit like 78.70 or 0.08).

Option: 1

0.4


Option: 2

0.81


Option: 3

1.23


Option: 4

-0.4


Answers (1)

best_answer

Cell reaction

Cathode: H_{2}O(l)+\frac{1}{2}O_{2}(g)+2e^{-}\rightarrow 2OH^{-}(aq)

Anode   :  H_{2}(g)\rightarrow 2H^{+}(aq)+2e^{-}

________________________________________________________________

              H_{2}O(l)+\frac{1}{2}O_{2}(g)+H_{2}(g)\rightarrow 2H^{+}(aq)+2OH^{-}(aq)

 

Also we have

H_{2}(g) +\frac{1}{2}O_{2}\rightarrow H_{2}O(l)                        \Delta G^{0}_{f} = -237.2 kJ/ mole

H_{2}O(l)\rightarrow H^{+}(aq)+OH^{-}(aq)          \Delta G^{0} = 80kJ/mole

Hence for cell reaction

\Delta G^{0} = -77.20 kJ/mole

so, E^{0}= -\frac{\Delta G^{0}}{nF} = \frac{77200}{2\times 96500} = 0.40 V

Posted by

shivangi.bhatnagar

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