Calculating cell potential

[![enter image description here](https://i.stack.imgur.com/EBK9E.png)](https://i.stack.imgur.com/EBK9E.png)


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:


[![enter image description here](https://i.stack.imgur.com/tE74G.gif)](https://i.stack.imgur.com/tE74G.gif)


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? 


[![enter image description here](https://i.stack.imgur.com/BCG3g.jpg)](https://i.stack.imgur.com/BCG3g.jpg)




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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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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