MCQ
The passage of current liberates ${H_2}$ at cathode and $C{l_2}$ at anode. The solution is
- ACopper chloride in water
- ✓$NaCl$ in water
- C${H_2}S{O_4}$
- DWater
✓
Answer
Correct option: B.
$NaCl$ in water
b
(b) Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $N{a^ + }$
(b) Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $N{a^ + }$
Cathode: ${H_2}O + {e^ - } \to \frac{1}{2}{H_2} + O{H^ - }$
Anode: $C{l^ - } \to \frac{1}{2}C{l_2} + {e^ - }$.
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