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Set up an electrochemical cell where the cathode is the system in question B, $\ce{Fe^2+ -> Fe^3+}\ E = 0.68~\textrm{V}$, whereas the anode is a saturated calomel electrode ($0.244~\mathrm{V}$ relative to standard hydrogen electrode). Determine potential of this electrochemical cell.

\begin{align} \ce{Fe^2+ +e^- &-> Fe^3+} & E &=0.68~\mathrm{V}\\ \ce{2Hg &-> Hg^2+ +2e^- } & E &=0.244~\mathrm{V} \end{align}

Usually, I just do $\text{Cathode (Fe) - Anode (Hg)}$ and get the correct answer. However, the correct answer is listed as $1.16~\mathrm{V}$. There are only two numbers to manipulate. I found that $$E(\ce{Fe}) + 2 E(\ce{Hg}) = 0.68~\mathrm{V} + 2\cdot 0.244~\mathrm{V} = 1.168~\mathrm{V}$$

Why is this? The amount of substance is not supposed to effect the electro-potential.

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    $\begingroup$ Please check your equations. Calomel is $\ce{Hg2Cl2}$! The equation should be $\ce{2Hg <=>Hg2^{2+} +2e-}$. $\endgroup$ – Klaus-Dieter Warzecha Mar 17 '16 at 8:41
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First of all, you have to use reduction potentials as IUPAC considers the reduction potentials to be called the standard potentials. It is not clear in your question whether you have used reduction or oxidation potentials.

Secondly, your equations don't make sense. The first equation shows Fe in +2 state getting reduced to Fe in +3 state. Your second equation shows oxidation of mercury from pure state to +1 state.

And you are right that stoichiometry should not affect the EMF of cell.

The correct cell reactions will be iron(II) getting oxidised to iron(III) and Mercury(I) being reduced to Mercury(0).

At Cathode, Reduction of mercury takes place.

At Anode , Oxidation of Iron takes place. But we always use only reduction potential to calculate EMF. Don’t use oxidation potential of one with reduction potential of the other.

And yes EMF= E(Cathode) - E(Anode) provided both are reduction potentials.

But with the given electrode potentials I see no way to get 1.16V as an answer for the EMF of cell.

I would suggest that you check the question and electrode potentials again. And check the answer key again. Maybe there is an error there. It would be best to check about that with your course instructor.

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    $\begingroup$ See, it's calomel electrode. There is no mercury (II). $\endgroup$ – Ivan Neretin Mar 17 '16 at 7:38
  • $\begingroup$ I see. I didn't notice that for some reason. I only saw the equation with Hg(II) and assumed that was correct. It must be Hg(I) not Hg(II). $\endgroup$ – shre_sudh_97 Mar 17 '16 at 15:29

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