The redox reaction can be split as

          $\begin{array}{cccc}{\left[\stackrel{+5}{\text{Cl}}{\text{O}}_3\right]}^{-1}& ⟶& {\left[\stackrel{+7}{\text{Cl}}{\text{O}}_4\right]}^{-1}+2\text{e}& \left(\text{OHR}\right)-\text{anode}\\ {\left[\stackrel{+5}{\text{Cl}}{\text{O}}_3\right]}^{-1}+2\text{e}& ⟶& {\left[\stackrel{+3}{\text{Cl}}{\text{O}}_2\right]}^{-1}& \left(\text{RHR}\right)-\text{cathode}\end{array}$

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       $\begin{array}{ccccc}2{\text{ClO}}_3^{-1}& \stackrel{{\mathrm{n\; =\; 2e}}^{-}}{\rightleftarrows }& {\text{ClO}}_4^{-1}& +& {\text{ClO}}_2^{-1}\end{array}$

${\text{E}}_{\text{cell}}^{\text{o}}={\text{E}}_{\text{red }\left(\text{cath}\right)}^{\text{o}}-{\text{E}}_{\text{red }\left(\text{anode}\right)}^{\text{o}}$

        = 0.33 - 0.36 = -0.03 V

$\mathrm{as }{\text{E}}_{\text{red }\left(\text{Anode}\right)}^{\text{o}}={\text{E}}_{{\text{ClO}}_4^{-1}/{\text{ClO}}_3^{-1}}^{\text{o}}=+0\text{.}36\text{ V}$

      $\left(=-{\text{E}}_{{\text{ClO}}_3^{-1}/{\text{ClO}}_4^{-1}}^{\text{o}}\right)$

at equilibrium E_cell = 0

${\text{E}}_{\text{cell}}^{\text{o}}=\frac{0\text{.}059}{\text{n}}\text{log }{\text{K}}_{\text{eq}}=\frac{0\text{.}059}{2}\text{log }{\text{K}}_{\text{eq}}$

Writing concentration of speces at equilibrium

$\begin{array}{ccccc}2{\text{ClO}}_3^{-1}& \rightleftharpoons & {\text{ClO}}_4^{-1}& +& {\text{ClO}}_2^{-1}\\ \left(1-2\text{x}\right)& & \left(\text{x}\right)& & \left(\text{x}\right)\\ \text{Molar}& & \text{Molar}& & \text{Molar}\end{array}$

${\text{E}}_{\text{cell}}^{\text{o}}=-0\text{.}03=\frac{0\text{.}059}{2}\text{ log }\frac{{\text{x}}^2}{{\left(1-2\text{x}\right)}^2}=0\text{.}059\text{ log }\frac{\text{x}}{1-2\text{x}}$

$\frac{-0\text{.}03}{0\text{.}059}=\text{log}\frac{\text{x}}{1-2\text{x}}\Rightarrow \text{log}\frac{1-2\text{x}}{\text{x}}=\frac{0\text{.}03}{0\text{.}059}$

$\text{log}\frac{1-2\text{x}}{\text{x}}=0\text{.}509$

$\frac{1-2\text{x}}{\text{x}}=3\text{.}229$

1 - 2x = 3.229 x $\Rightarrow$ 5.229x = 1

$\text{x}=\frac{1}{5\text{.}229}=0\text{.}191\text{ M}$.