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What is $E^\circ$ for the half-reaction representing the oxidation of $\ce{HOCl}$ to $\ce{ClO3-}$ given the equations below? \begin{align} \ce{ClO3- +6 e− + 6 H+ &= Cl− +3 H2O},& E^\circ &= \pu{1.446 V}\\ \ce{HOCl + 2 e− + H+ &= Cl− + H2O},& E^\circ &= \pu{1.481 V}\\ \end{align}

It seems too simple, but would the answer be to take $$E^\circ = \pu{1.481 V} - (\pu{-1.446 V}) = \pu{2.927 V}?$$

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For adding up equations, you can't directly add $E^0$,unless you know that the no. of electronic transfres are same in both the reactions given.So, it is safe to calculate by $\Delta$$G^0$ calculation,
For $HOCl$ reaction , you have,
$HOCl + 2e^- + H^+ = Cl^- + H_2O$ -- $\Delta$$G^0$= -$nFE^0$ = $-2*F*1.481$
By inverting the First reaction, sign of $E^0$ will be changed and thus,
$Cl^- + 3H_2O = ClO_3^- + 6e^- + 6H^+$ --- $\Delta$$G^0$=-$nFE^0$ = $-6*F*(-1.446)$
Adding the two reaction ,
$HOCl + 2H_2O = ClO_3^- + 4e^- + 5H^+$ -- $\Delta$$G^0$= - $(4*F*E^o)$
$\Delta$$G^0$ will be added and for the whole reaction,
$E^0(total) $ =
$\Delta$$G^0$(total)/$nF$= $-F*(2*1.481-6*1.446)/(-4*F)$)=$-1.4285$ $V$.
So, $E^0$ for the oxidation from $HOCl$ to $ClO_3^-$ is $-1.4285$ $ V$.
Thus, the answer is different from simply adding the $E^0$ for reactions which is not correct for all the cases(is correct when electron transfer is same for both reactions).

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