Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (b) Ni1s2 + 2 Ce4+1aq2 ¡ Ni2+1aq2 + 2 Ce3+1aq2

Video transcript

Hello everyone. So in this video we're given the half cell reactions as well as their respective standard production potentials over here for the overall reaction right over here. Alright, so we want to calculate for the standard cell potential for this overall reaction. So we can kind of break down this whole concept first. So for our cathode this is where reduction occurs. The reduction is when the oxidation number decreases. And then for our A node is where the oxidation occurs. An oxidation is when the oxidation number increases. So with our balanced reaction or overall reaction that we have going on here, we kind of evaluate this. So we can see here for our first starting material of H. C. L. O. which we have six moles of The off station number of hydrogen with a non metal is positive one. So again, H has a positive one value. And then for ox oxygen is always negative too. That leaves us with the oxidation number of chlorine to be positive one. To kind of balance out everything. Now for our second reactant of age plus of course we have that plus one charge. As automation number will be positive one. Okay, now for the third Regent we have a U. So it's of course we see that A. U. Is in its natural mental state. So the off station number will be of course zero. All right. Now looking at the products. So these are the reactant. Now looking at the products we have the C. L. Two. Of course that's a di atomic molecules. So it's an inter natural elemental state. Therefore the oxidation number is zero. Then we have our H20. So oxidation number of hydrogen with a non metal is positive one and the oxidation number of oxygen is always negative too. Last C A. U. Three plus has a plus three charge. So its oxidation number will also be plus three. Alright, so I'm looking at each individual adam, we see that there's an oxidation change or oxygen state change of our chlorine in hcl O. Going to cl two guests. So cl is going from the oxidation state Going from positive 1-0. So because there's a decrease is being reduced and this happens at the cathode. Now for our au We see that it's going from off station state of zero two positive three. So the oxidation number has increased, meaning that it's been oxidized and this occurs at the node. All right now, cal going for R. S. L. Value. No, the valley is just equal to the center reduction potential of the cathode minus the unload. And of course, were given its value already and the problem will just go ahead and plug those in. So we have 1.63V -1.498V. Giving us the sl value of 0.13V. So that is going to be my final answer for this problem. Thank you all so much for watching