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General Principles and processes of Isolation of Elements
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A solution of $Na_2SO_4$ in water is electrolysed using inert electrodes. The products at cathode and anode are respectively
A
$O_2\, ; \, H_2$
B
$O_2 \, ; \, Na$
C
$H_2 \, ; \, O_2$
D
$O_2\, ; \,SO_2$
Solution
According to the standard electrochemical series, elements present above in the series get reduced easily and present lower in the series gets oxidized more easily.
$\therefore$ The reduction potentials of $H ^{+}=0, Na ^{+}=-2.7$.
$\therefore H ^{+}$ gets reduced and $OH ^{-}$gets oxidised.
At cathode: $2 H ^{+}+2 e ^{-} \rightarrow H _2$
At anode: $2 OH ^{-} \rightarrow H _2 O +\frac{1}{2} O _2+2 e ^{-}$
Standard 12
Chemistry
Similar Questions
Match List$-I$ with List$-II$
| List$-I$ | List$-II$ |
| $(a)$ Sodium Carbonate |
$(i)$ Deacon |
| $(b)$ Titanium | $(ii)$ Castner-Kellner |
| $(c)$ Chlorine | $(iii)$ Van-Arkel |
| $(d)$ Sodium hydroxide | $(iv)$ Solvay |
Choose the correct answer from the options given below :
