2
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.