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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
$O_2,\,H_2$
$O_2,\,Na$
$H_2,\,O_2$
$O_2,\,SO_2$
Solution
A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are $\mathrm{H}_{2}, \mathrm{O}_{2}$ respectively. Hydrogen ions having lower discharge (or higher reduction) potential than sodium ions will be liberated at the cathode.
$2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \uparrow$
Hydroxide ions having a lower discharge potential than sulphate ions, will be liberated at anode.
$4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 \mathrm{e}^{-}$
Similar Questions
Match List $- I$ with List $- II.$
| List $-I$ (Colloid Preparation Method) | List $-II$ (Chemical Reaction) |
| $(a)$ Hydrolysis | $(i)\;2 \mathrm{AuCl}_{3}+3 \mathrm{HCHO}+3 \mathrm{H}_{2} \mathrm{O}\rightarrow\;{2 \mathrm{Au}(\mathrm{sol})+3 \mathrm{HCOOH}+} {6 \mathrm{HCl}}$ |
| $(b)$ Reduction | $(ii)\;\mathrm{As}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{As}_{2} \mathrm{S}_{3}(\mathrm{sol})+3 \mathrm{H}_{2} \mathrm{O}$ |
| $(c)$ Oxidation | $(iii)\;\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{S} \rightarrow 3 \mathrm{S}(\mathrm{sol})+2 \mathrm{H}_{2} \mathrm{O}$ |
| $(d)$ Double Decomposition | $(iv)\;\mathrm{FeCl}_{3}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{sol})+3 \mathrm{HCl}$ |
Choose the most appropriate answer from the options given below.
