How to calculate the electrochemical potential of a cell using known half-cell potentials?


> 
> 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 
$$E^\circ = \pu{1.481 V} - (\pu{-1.446 V}) = \pu{2.927 V}?$$




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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