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First of all, is the problem even worded correctly? From how the problem is worded, it seems that copper is being oxidized as well as iodide ion. This is what I am taught:

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Second of all, I'm not getting 0.24 volts as the cell potential. I'm getting 0.19 volts.

Here is my work. What am I doing wrong?

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    $\begingroup$ I think you are almost there: log(.009)~-2 $\endgroup$ – Dan Burden Jul 21 '16 at 0:35
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It's important to note that just subtracting the cell potentials only work if everything is at 1M, but in the question you've got slightly different concentrations. So actually all you've missed the last step! Your calculations are all correct. You've stopped at:

$$E_{cell} = 0.19 - \frac{0.0592}{2}\log{0.009}$$

and if you evaluate this you'll find that:

$$E_{cell} = 0.19 + 0.060 = 0.25 V$$

Which is just a little bit off from 0.24V

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  • $\begingroup$ oh wow! To think that 2 hours of frustration was due to a simple calculator error ... $\endgroup$ – Dissenter Jul 21 '16 at 1:19
  • $\begingroup$ why would the problem list the second half-reaction backwards though? $\endgroup$ – Dissenter Jul 21 '16 at 1:22
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In electrochemistry you always write the half reaction with reduced species in the left and oxidized species on the right. This is done for consistency purposes. Species with higher potential will reduce and species with lower potential will be oxidize.

I$_2$ + 2e- ----> 2 I$^-$ +0.54 //will reduce (go from left to right), has positive charge
Cu$^{2+}$ + 2e- ---> Cu +0.34 //will oxidize(go from right to left), has negative charge

Next, you should use Nernst equation to each half reaction. You will get values E$_{Cu^{2+}/Cu}$ and E$_{I_2/I^{2+}}$. Now subtract one from another to get the voltage.

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  • $\begingroup$ I thought in cell notation the reactant is on the left. $\endgroup$ – Dissenter Jul 21 '16 at 0:49
  • $\begingroup$ @Dissenter the oxidized form is on the left. Here is why: Cu/Cu$^{2+}$ is reduced by zinc, but is oxidized by Ag/Ag$^{+}$. It is impractical to have two tables (one for each case), so scientists agreed to always write the reaction in the direction of reduction. $\endgroup$ – sixtytrees Jul 21 '16 at 0:55
  • $\begingroup$ what do you mean the copper is reduced by zinc and oxidized by silver? There is no zinc or silver. $\endgroup$ – Dissenter Jul 21 '16 at 0:59

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