Redox Reactions And Electrochemistry Ques 248

Question: The passage of current liberates $ H_2 $ at cathode and $ Cl_2 $ at anode. The solution is [EAMCET 1979,87]

Options:

A) Copper chloride in water

B) $ NaCl $ in water

C) $ H_2SO_4 $

D) Water

Show Answer

Answer:

Correct Answer: B

Solution:

Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $ N{a^{+}} $ Cathode: $ H_2O+{e^{-}}\to \frac{1}{2}H_2+O{H^{-}} $ Anode: $ C{l^{-}}\to \frac{1}{2}Cl_2+{e^{-}} $ .



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