Redox Reactions And Electrochemistry Ques 615
Question: At 298K the standard free energy of formation of $ H_2O(l) $ is $ -237.20kJ/mol $ while that of its ionisation into $ {H^{+}} $ ion and hydroxyl ions is 80 kJ/ mol, then the emf of the following cell at 298 K will be [Take Faraday constant F = 96500 C] $ H_2(g,1bar)|{H^{+}}(1M)||O{H^{-}}(1M)|O_2(g,1,bar) $
Options:
A) 0.40 V
B) 0.81 V
C) 1.23 V
D) $ -0.40V $
Show Answer
Answer:
Correct Answer: D
Solution:
[a] Cell reaction cathode: $ \begin{aligned} & H_2O(l)+\frac{1}{2}O_2(g)+2{e^{-}}\xrightarrow{{}}2O{H^{-}}(aq) \\ & ,anode:H_2(g)\xrightarrow{{}}2{H^{+}}(aq)+2{e^{-}} \\ & \overline{H_2O(l)+\frac{1}{2}O_2(g)+H_2(g)\xrightarrow{{}}2{H^{+}}(aq)+2O{H^{-}}(aq)} \\ \end{aligned} $ Also we have $ H_2(g)+\frac{1}{2}O_2(g)\xrightarrow{{}}H_2O(l); $ $ \Delta G_f^{{}^\circ }=-237.2kJ/mole $ $ H_2O(l)\xrightarrow{{}}{H^{+}}(aq)+O{H^{-}}(aq); $ $ \Delta G{}^\circ =80kJ/mol $ Hence for cell reaction $ \Delta G{}^\circ =-237.2+( 2\times 80 )=-77.20,kJ/mol $
$ \therefore E{}^\circ =-\frac{\Delta G{}^\circ }{nF}=\frac{77200}{2\times 96500}=-0.40V $
