Consider the following substances: Fe(s), PbO2(s), H+(aq), Al(s), Ag(s), Cr2O72-(aq). (d) Which substances can be oxidized by Cu2+(aq)? Which can be reduced by H2O2(aq)?

Video transcript

Hello. Everyone in this video, we're being asked to consider these substances for our two questions right over here, A and B. So the values that I'm going to refer to are found in the table of standard reduction potentials. So this table can be either given to you by a professor or found in your textbook. So I'm just referring to this table of standard reduction potentials for each of our substances. So starting off with our each C L. O. That has A. S. L. Value of 1.61V and then our I. 03 negative one. Um Power here, this is Aquarius and the resell value for this is 1. volts. Next Substances are HN 02. This has the value of 0.98V. Then we have our B I. Which they solid. This sl value. Is there a .32V. Next is our H two C 204. This has a value of -0.49V. And lastly our zinc, which is a solid. Its value is equal to negative 0.76V. Alright, so for question A Then it says H G two cl two solid is capable of oxidizing which of these following substances. So referring again to my table of standard reduction potentials, we can see here that the reaction for H g two cl two joe. Go ahead and write out we're adding two electrons. This to give us two moles of our mercury. That's a liquid state and two moles of chlorine and islands where the cell value then, is there a 20.28 volts. So because our molecule of interest, so the HG two cl 2 as a value of zero points to 8V. It's an oxidizing agent. So I can oxidize oxidize substances with a lower e cell value. Can oxidize. So let's write out those substances. We can see here. Then looking at These values here in green and comparing it to this 0.28V, we can see that our substance of interest can go ahead and subsidy or oxidize our H two C 204 as well as our solid sink. So that's the answer for a. Now let's go ahead and do B. So again, looking at my table center reduction potentials, The compound of interest now as H three a. s. 03. Which is a curious reacting that with two moles of our hydrogen Canadians as well as two electrons. And that yields H three A. S. 03, one mole of liquid water. And the sl value for this Is equal to 0.56V. So because our molecule interest, so the H three A. S. 03, That has a value that's equal to 0.56V. This is a reducing agents and this can reduce substances with a higher value. So then this can reduce again, we're just comparing the values that we have in green here to this value here and therefore we can see that this molecule interest can reduce. H c l O I owe 3 -3 as well as each HNO two. Alright, so this is the final answer for part B of this problem. Thank you all so much for watching.