The following galvanic cell has a potential of 1.214 V at 25 °C: $$\mathrm{Hg}_{(l)}|\mathrm{Hg}_{2}\mathrm{Br}_{2(s)}|\mathrm{Br}^-(0.10\,\mathrm{M})||\mathrm{MnO}_{4}^-(0.10\,\mathrm{M})|\mathrm{Mn}^{2+}(0.10\,\mathrm{M}),\mathrm{H}^+(0.10\,\mathrm{M})|\mathrm{Pt}_{(s)}$$ Calculate the value of K sp for Hg 2 Br 2 at 25 °C.

The half reactions are

$\mathrm{anode}: \,\,\,5\times[2\,\mathrm{Hg}_{(l)}+2\,\mathrm{Br}^-_{(aq)}\rightarrow \mathrm{Hg}_{2}\mathrm{Br}_{2(s)}+2\,e^-]$

$\mathrm{cathode}:\,\,\,\,2\times[\mathrm{MnO}_{4(aq)}^- + 8\,\mathrm{H}^+_{(aq)}+5\,e^-\rightarrow \mathrm{Mn}^{2+}_{(aq)}+4\,\mathrm{H}_{2}\mathrm{O}_{(l)}]$

$\mathrm{overall}:\,\,\,2\,\mathrm{MnO}_{4(aq)}^-+10\,\mathrm{Hg}_{(l)}+10\,\mathrm{Br}^-_{(s)}+16\,\mathrm{H}^+_{(aq)}\rightarrow 2\,\mathrm{Mn}^{2+}_{(aq)}+5\,\mathrm{Hg}_{2}\mathrm{Br}_{2(s)}+8\,\mathrm{H}_{2}\mathrm{O}_{(l)}$

The cell potential of the anode is $E1=-0.142 \,V$. However, after determining this, the solution manual does this.

$\mathrm{overall}:\,\,\, 2\,\mathrm{Hg}_{(l)}\rightarrow \mathrm{Hg}_{2(aq)}^{2+}+2\,e^-$

$\mathrm{reduction}:\,\,\, \mathrm{Hg}_{2}\mathrm{Br}_{2(s)}+2\,e^-\rightarrow 2\,\mathrm{Hg}_{(l)}+2\,\mathrm{Br}^-_{(aq)}$

$\mathrm{overall}:\,\,\,\mathrm{Hg}_{2}\mathrm{Br}_{2(s)}\rightarrow \mathrm{Hg}_{2(aq)}^{2+}+2\,\mathrm{Br}^-_{(aq)}$


After determining E2, it uses that value to determine the equilibrium expression. My question is, why can we not use E1?


This question is very confusing not least because the numbers just don't add up.

The anode voltage (i.e. the voltage requirement for the oxidation of $\ce{Hg}$) is given in the question as $-0.142 \, \mathrm{V}$ whilst the tabulated version is $-0.80 \, \mathrm{V}$ (although it could be lower since mercury dibromide is less soluble than the theorised $\ce{Hg2^2+(aq)}$ in the formal tables). The cathode potential is also odd because the tabulated version is definitely $1.51 \, \mathrm{V}$ and not the voltage you've given, which is disturbingly negative.

I have no idea what is going on, but there's some mistake in here somewhere. I'll try and do it my way and describe everything I'm doing, and see if I can approximate a correct answer. I'm going to calculate my version of the standard cell potential.

As you might be aware, the line convention in electrochemistry places the anode reaction to the left of the salt bridge symbol (||) with the cathode reaction to the right. The anode is where oxidation occurs, and reduction at the cathode. Immediately we can see that when we look at the tabulated reduction potentials to work this out, we'll have to reverse our anode reaction because it isn't reduction, it's oxidation!

So looking first at our anode reaction, the reaction must involve $\ce{Hg (l)}$ and $\ce{Hg+(aq)}$. A reasonable reduction reaction relationship involving these two species can be found by a simple internet search. A good place to start is:

$$ \ce{Hg_2^2+(aq)} + \ce{2e-} \ce{->} \ce{2Hg(l)} $$

Which occurs with a potential difference of $\ce{0.80V}$. However as we've said, this reaction occurs at the cathode, so we have to reverse it, along with the voltage requirement:

$$ \ce{2Hg(l)} \ce{->} \ce{Hg_2^2+(aq)} + \ce{2e-} $$

A slight final modification is needed to this above equation, since our oxidation state of $\ce{Hg_2^2+(aq)}$ does not occur in isolation, but as the molecule $\ce{Hg_2Br_2(s)}$.Thus, just substituting molecules for imaginary ions:

$$ \ce{2Hg(l)} + \ce{2Br^-}\ce{->} \ce{Hg_2Br_2(s)} + \ce{2e-} $$

Notice that the addition of $\ce{Br^-}$ species does not change our oxidation voltage ($-0.80 \, \mathrm{V}$ since this is a spectator ion, and does not change oxidation number).

So the above is a valid reaction for what is occurring at the anode. We now do the same for the cathode. At the cathode, a reduction is happening, so we can use our reduction potentials directly without needing to reverse anything.

Thus we start with the simplest equation representing this process:

$$ \ce{Mn^7+} + \ce{5e-}\ce{->} \ce{Mn^2+} $$

And add in our actual molecules, being sure to balance by adding water and protons to each side as required.

$$ \ce{MnO4^-} + \ce{5e-} + \ce{8H^+} \ce{->} \ce{Mn^2+} + \ce{4H2O} $$

Which is our cathode reaction. Now we need to recombine both the cathode and anode reactions to give the overall scheme, but note that the cathode reaction requires 5 electrons, whilst the anode reaction only produces 2. Therefore we multiply both by a number which allows us to reach the lowest common denominator of each (10 in this case), i.e. a factor of 5 for the anode reaction and 2 for the cathode reaction.

So the overall equation is:

$$ \ce{2MnO4^-} + \ce{16H^+(aq)} + \ce{10Hg(l)} + \ce{10Br^-}\ce{->} \ce{2Mn^2+(aq)} + \ce{4H2O(l)} + \ce{5Hg_2Br_2(s)} $$

We add the individual half reaction voltages of $-0.80 \, \mathrm{V}$ and $+1.512 \, \mathrm{V}$ to give $\ce{E_cell}$ of $+0.712 \, \mathrm{V}$, which is curiously different from the one you gave. So far we've only calculated the standard cell potential, and not the actual cell potential using the non-standard concentrations in the Nernst equation, however this has been given to us as $1.214 \, \mathrm{V}$. Having an actual potential higher than the standard at lower than unit concentrations doesn't seem likely to me.

So I'm betting that a mistake in the question, or in your transcription of the question has contributed to this problem which I can't seem to solve. I'd be happy to hear others' responses to this, or a more direct transcription of the question. I'd be happy to help more once this comes to light.


From a quick look-see, I figure you can use $E_{1}$, but you have to use it in a slightly modified form.

Note that $E_{2}$ is only a modification of $E_{1}$ which instead of using $Hg_{(l)}$, uses $Hg^{2+}$, which is the appropriate form for calculating the $K_{sp}$ with.

To change $E_{1}$ to $E_{2}$, a substitution of $Hg^{2+} + 2e^{-} \rightarrow Hg_{(l)}$ is necessary.

After rereading, I notice that your question might not exactly want what I'm answering. An alternative explanation is that $Hg_{2}Br_{2}$ does not dissociate into $Hg_{(l)}$ in solution, but $Hg^{2+}$ which is why using the ion is necessary.

  • 2
    $\begingroup$ This answer doesn't actually make any sense, what do you mean by 'converting E1 into E2?' they are two completely separate voltages which are dependent on the specific half-reaction. $\endgroup$ – user7232 Aug 31 '14 at 16:30
  • $\begingroup$ Sorry if was not clear, I meant the half-reactions associated with voltages E1 and E2. $\endgroup$ – Richard Aug 31 '14 at 17:02

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