A) \[{{H}_{2}}\] is anode and \[Cu\] is cathode
B) \[{{H}_{2}}\] is cathode and \[Cu\] is anode
C) reduction occurs at \[{{H}_{2}}\] electrode
D) oxidation occurs at \[Cu\] electrode.
Correct Answer: A
Solution :
As Cu is .below hydrogen in the electrochemical series. Hydrogen is oxidised at anode and comes out as \[{{H}_{2}}\] while Cu is reduced at the cathode. \[2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}},{{E}^{o}}=0.0volt\] \[C{{u}^{2+}}+2{{e}^{-}}\to Cu,\,{{E}^{o}}=0.37volt\]
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