When an aqueous solution of sodium sulfate (N | vedclass

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(A) The dissociation of sodium sulfate in water is: $Na _2 SO _4(a q) \rightarrow 2 Na ^{+}(a q)+ SO _4^{2-}(a q)$
Water also dissociates slightly: $

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$H_2, O_2$

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(A) The dissociation of sodium sulfate in water is: $Na _2 SO _4(a q) \rightarrow 2 Na ^{+}(a q)+ SO _4^{2-}(a q)$
Water also dissociates slightly: $H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}$.
At the cathode,$H^+$ ions are reduced in preference to $Na^+$ ions because the reduction potential of $H^+$ is higher: $2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}$.
At the anode,$OH^-$ ions are oxidized in preference to $SO_4^{2-}$ ions: $4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-$.
Therefore,the products at the cathode and anode are $H_2$ and $O_2$,respectively.

When an aqueous solution of sodium sulfate $(Na_2SO_4)$ is electrolyzed using platinum electrodes,what are the products at the cathode and anode,respectively?

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