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(C) In an aqueous solution of $NaBr$,the ions present are $Na^+$,$Br^-$,$H^+$,and $OH^-$. At the cathode,the reduction potential of $H_2O$ $(-0.83 \ V

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$H_2, Br_2$ and $NaOH$

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(C) In an aqueous solution of $NaBr$,the ions present are $Na^+$,$Br^-$,$H^+$,and $OH^-$. At the cathode,the reduction potential of $H_2O$ $(-0.83 \ V)$ is higher than that of $Na^+$ $(-2.71 \ V)$,so $H_2$ gas is evolved: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$. At the anode,the oxidation potential of $Br^-$ is higher than that of $H_2O$,so $Br_2$ is produced: $2Br^-(aq) \rightarrow Br_2(l) + 2e^-$. Consequently,$H_2$ gas is produced at the cathode,$Br_2$ is produced at the anode,and $NaOH$ remains in the solution.

During the electrolysis of an aqueous solution of $NaBr$ using inert electrodes,what are the products formed?

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