During the electrolysis of concentrated H_2SO | vedclass

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(A) At the anode,the following reactions occur: $2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-$ $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$ Let the moles of

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thrice that of $O_2$ in moles

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(A) At the anode,the following reactions occur: $2H_2SO_4 ightarrow H_2S_2O_8 + 2H^+ + 2e^-$ $2H_2O ightarrow O_2 + 4H^+ + 4e^-$ Let the moles of $H_2S_2O_8$ be $x$ and the moles of $O_2$ be $x$ (since they are equimolar). Total moles of electrons lost at the anode = $(2 	imes x) + (4 	imes x) = 6x$. At the cathode,the reduction reaction is: $2H^+ + 2e^- ightarrow H_2$ Total moles of electrons gained at the cathode must equal the total moles of electrons lost at the anode,which is $6x$. Moles of $H_2$ produced = $6x / 2 = 3x$. Since $n(O_2) = x$,the moles of $H_2$ produced is $3 	imes n(O_2)$.

During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed in equimolar amounts. The amount of $H_2$ that will be formed simultaneously is $(2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-)$

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