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^{-}} $ .
