Question #44126

An electrochemical cell contains $1.0\text{ L}$ of an aqueous solution comprising $2.0\text{ M } \ce{NaCl}$ and $0.1\text{ M } \ce{Na2SO4}$ at $298\text{ K}$. Electrolysis is performed using inert platinum electrodes by passing a constant current of $9.65\text{ A}$ for $10,000\text{ s}$. The primary gas evolved at the anode is collected and completely reacted with an excess of hot, concentrated $\ce{NaOH}(aq)$.

Assuming $100\%$ current efficiency for the kinetically favored anodic process, identify the major chlorine-containing oxyanion formed in the final solution and calculate its amount.

(Given: $1\text{ F} = 96500\text{ C mol}^{-1}$; Standard oxidation potentials: $E^\circ_{\ce{Cl-}/\ce{Cl2}} = -1.36\text{ V}$, $E^\circ_{\ce{H2O}/\ce{O2}} = -1.23\text{ V}$, $E^\circ_{\ce{SO4^{2-}}/\ce{S2O8^{2-}}} = -2.01\text{ V}$)