Finding voltage through values in an appendix

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To find the Ered° values for the cathode and anode in the given reaction, the standard potentials for Ce4+/Ce3+ and the reverse reaction for Bi(s) must be utilized. The reverse reaction's voltage can be used directly without flipping the sign, as cell voltage is determined by the difference between standard potentials. It is important to remember that standard potentials are conventionally listed as reduction potentials, but the actual reaction direction will depend on the specific half-cells involved. Therefore, one can simply calculate the cell voltage by taking the difference between the relevant standard potentials. Understanding this concept eliminates the need to adjust the sign for the reverse reaction.
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Homework Statement



Find the Ered° values in the appendix for the cathode and anode in the reaction

3Ce4+(aq) + Bi(s) + H2O(l) → 3Ce3+(aq) + BiO+(aq) + 2H+(aq)

Homework Equations



None.

The Attempt at a Solution



So in my appendix there is a value for the Ce4+/Ce3+ half-reaction, but only the reverse reaction for Bi(s) + H2O(l) → BiO+(aq) + 2H+(aq). What do you do then?
 
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Hi silenzer. I'd guess that you simply take the negative of that voltage. But it's a long time since I did any chem. My recommended approach is to do a web search for a similar worked example to see whether that's the way. I'd be surprised if your textbook didn't have a worked example.

Of course, if you know the correct answer for your question, you can work it out and compare to see whether my guess is right.
 
While tables of standard potentials are often called "reduction potential tables", and reactions are written as reduction, it is just a matter of convention (makes it easier to find what you are looking for). Reaction goes both ways at a given potential. Or more precisely, when you mix all substances in their standard states (1 bar, 1M), and you put an inert electrode into the solution, you will read the standard potential as given. Whether the observed reaction will be reduction or oxidation will depend solely on the second half cell.

In other words - no need to flip the sign, cell voltage is just a matter of distance between standard potentials, regardless of whether they are given as reduction or oxidation.
 
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