Here it says given the follow redox reaction we have cerium solid plus aluminum ion reacting to produce aluminum solid plus cerium aqueous. Here it says find its standard sub potential when given the following half reactions. Here we're given the half reaction for the cerium 3 plus ion and the aluminum ion.
Now here the steps that we take is first refer to the redox reaction to determine which species is reduced and which is oxidized. This is incredibly important because remember that the cathode is a site of reduction, so the cathode equals reduction and the anode is a site of oxidation, so the anode equals oxidation. Once we know this, we can use the standard subpotential formula to find our final answer. And that's because standard subpotential, so Ecell, equals cathode minus anode.
If we take a look here, we see that cerium solid goes from zero for its oxidation state to +3. Its oxidation number increased, therefore it has been oxidized. So here, if it's the site of oxidation, that means it must be the anode. And then if we look at aluminum, aluminum goes from what? It goes from plus three in terms of its charge/oxidation state to neutral to its natural standard state, which is 0. So it goes from plus three to zero. Its oxidation number has been reduced or decreased, so reduction has occurred. Therefore it is the cathode.
So we've determined the cathode and the anode, and now we can find the standard cell potential. So take the values and plug them in. So for this will be -1.677 - (-2.336). Remember, a minus of a minus just means that you're adding them together. So -1.677 + 2.336 gives me at the end 0.659 volts. This will represent the answer from my standard cell potential for the following example question.