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When pure $H_2SO_4$ is electrolysed,the product at the anode is

$\begin{array}{1 1}H_2S_2O_7\\H_2S_2O_8\\H_2S_2O_3\\H_2S_4O_6\end{array}$

Can you answer this question?

Answer : $H_2S_2O_8$
$2H_2SO_4\rightarrow H_2S_2O_8+2H^++2e^-$
answered Jul 24, 2014