Determining Overall Equation and Oxidation Potential


> 
> Cell 1 is composed of a reference chloranil half-cell (standardised against SHE as $+0.680 \mathrm{V}$) connected to an $\ce{Ag/AgCl}$ half-cell containing saturated ($0.125\ \mathrm{mol\ L^{−1}}$) $\ce{NaCl}$ in glacial acetic acid.
> 
> 
> $\ce{Ag\_{(s)}\ |\ AgCl\_{(s)}\ |\ Cl^{-}\_{(CH3COOH)}\ ($0.125$~M)\ ||\ H+(aq)\ ($0.5$~M),\ C6Cl4O2\_{(aq)(sat)},C6Cl4(OH)2\_{(aq)(sat)}|\ Pt\_{(s)}}$
> 
> 
> Cell 2 is composed of the same reference chloranil half-cell connected to an Ag/AgCl half-cell containing $\mathrm{1.0 ~mol~ L^{-1}}$ $\ce{HCl}$ in glacial acetic acid.
> 
> 
> $\ce{Ag\_{(s)}\ |\ AgCl\_{(s)}\ |\ HCl\_{(CH3COOH)}\ ($1.0$~M)\ ||\ H+(aq)\ ($0.5$~M),\ C6Cl4O2\_{(aq)(sat)},C6Cl4(OH)2\_{(aq)(sat)}|\ Pt\_{(s)}}$
> 
> 
> **a)** Write the equation for the overall reaction occurring in both cells.
> 
> 
> Cell 1 is used to determine the value of $\mathrm{E^o}$ for the reduction of $\ce{AgCl\_{(s)}}$ to $\ce{Ag\_{(s)}}$ and $\ce{Cl^{–}}$ in glacial acetic acid –
>  $\mathrm{E^\circ}\ce{(AgCl/Ag)}$. The cell potential of Cell is $0.422\ \mathrm{V}$
> 
> 
> **b)** Calculate $\mathrm{E\_{OX}}$ – the reduction potential of the oxidation half-cell – of Cell 1.
> 
> 
> 


**My Attempt**


**a)** Electro chemistry is not my strong point and I am not too sure on the notation for the cell. I have never seen a cell that has 3 species in the anode and in the cathode half-cell. Could someone explain to me what it means?


I am guessing the oxidation half-equation is: $$\ce{Ag\_{(s)} + Cl^{-}\_{(aq)} -> AgCl\_{(s)} + e^{-}}$$
My overall equation is then:
$$\ce{2Ag\_{(s)} + 2Cl^{-}\_{(aq)} + C6Cl4O2\_{(aq)} + 2H^{+} -> 2AgCl\_{(s)} + C6Cl4(OH)2\_{(aq)}}$$


Is this correct? Also will the overall reaction for cell 2 be the same as the overall reaction for cell 1 as the only difference is that $\ce{NaCl}$ is replaced with $\ce{HCl}$.


**b)** Is the question a typo and really means: "Calculate $\mathrm{E\_{ red}}$ – the reduction potential of the oxidation half-cell – of Cell 1." or am I missing something?


If the question is a typo, I get that the answer is $-0.258\ \mathrm{V}$, is this correct?


