The standard reduction potentials of the following halfreactions are given in Appendix E: Ag+1aq2 + e- ¡ Ag1s2 Cu2+1aq2 + 2 e- ¡ Cu1s2 Ni2+1aq2 + 2 e- ¡ Ni1s2 Cr3+1aq2 + 3 e- ¡ Cr1s2 (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value.

Video transcript

Hello everyone. So in this video we're given all these half reactions along with the respective centered production potentials right over here on the right side. So we're trying to identify the combination of these half cell reactions that will result in an overall reaction with the smallest positive the cell value as well as that value of the cell. So we know that if we want the smallest positive value, want the two half reactions. Who standard reduction potential values are closest in magnitude and science will combine. So with that in mind, we can see here that B. N. C. Has that requirement. So we know that we have a lower reduction potential value. An oxidation occurs and of course the oxidation occurs at the node. Now on the opposite end, if we have a higher standard reduction potential, we know that this is going to be reduced and this happens at the cathode. So now dealing with R B and C half equations we see for B We have a value of 0.913, so have a more negative value. Therefore it's going to be oxidized or this oxidation occurs there and occurs at the anodes and by default then the other will be at the cathode. Alright, let's go ahead and look at these two half cell reactions. So I'm just gonna go ahead and rewrite them now for our equations, what I'm gonna do is flip this and that will produce let's do this in different colors. Doing green. It will turn to Let's see so we have zinc two plus to give us a solid state of zinc. And then since we switch, switch this, the product now becomes the reactant and the reactors become the products. So we see now that the two electrons will go ahead and cancel. So the overall reaction is zinc two plus, which is Aquarius, and the chromium solid To give us c. n. solid and cr two plus Aquarius. So with that reaction in mind, let's go ahead and calculate for sl of course those standard reduction potentials are given to us in the problem. So it's of the cathode minus the standard reduction potential of a node. So that just equals to negative 0.760 volts minus the negative 0.913 volts. Putting this into the calculator, begins. The sl Is equal to 0.15 thoughts. So this is going to be my final answer for this problem. Thank you all so much for watching