<p>In order for an electrolytic cell to be able to reverse a reaction, doesn’t the applied voltage have to be greater than the voltage of the forward reaction?</p>
<p>Because in the Princeton Review book, one of the problems gives 3 possible reactions and their reduction potentials. Then it asks which 2 would occur given an applied voltage of 2 volts in an electrolytic cell.</p>
<p>The reasoning they use in picking the 2 reactions is that since those 2 are the only ones whose negative voltages are less than 2 volts. However, when you add the 2 reactions together, you get a total voltage of 2.2 V > 2.0 volts. So is Princeton Review wrong?</p>
<p>Are the three possibilities half-reactions or regular reactions? Half-reactions need to be added (making sure one is written as a reduction and one as an oxidations) and then the applied volatage must be greater than the sum of the half-reactions. However, if the three possibilities are full redox reactions, you wouldn’t combine them. Can you be mores specific (or tell me which PR edition and problem you are working on)?</p>
<p>Their reasoning is that reverse of last equation and first equation occur, at the anode and cathode respectively. However, if you add them, you get -2.2 V. The applied voltage is 2.0 V.</p>
<p>The problem is #13 pg. 233 of 2006-2007 PR AP Chemistry.</p>
<p>Sorry about the delay in responding. On the one hand, I checked an older version of PR and they use essentially the same problem (albeit on a different page) and use the same applied volatage of 2.0 V. You would think they would have corrected any typos between editions.</p>
<p>Nonetheless, it must be a typo. They correctly indicate that water is reduced preferentially rather than Na+ because it has the more positive reduction potential. The oxidation of Cl- is the only option they provide so that has to be the oxidation half-reaction, and they definately add to 2.2, so that would be the minimum applied voltage for electrolysis. Can’t explain why the error has persisted.</p>