You say you've looked up these values. My values won't be the same as yours because I'm using an old book but they will be close. Here is how you do the first one.
Cr==> Cr^3+ + 3e You look in your table and find this potential. It will be listed "backwards" as Cr^3+ + 3e ==> Cr because tables list them as reduction potentials and this is an oxidiation; my book lists it as -0.74 so we write 0.74 for Eo as written.
The second part of the reaction will be
Ca^2+ + 2e --> Ca and that is the way you look it up. My text lists that as -2.8 as a reduction.
We summarize
Cr ==> Cr^3+ + 3e Eo = 0.74
Ca^2+ + 2e ==> Ca Eo = -2.86
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So you add these two equations and obtain -2.12v and since it isn't a + voltage you know it won't go as written and now you know Ca^2+ ion will not oxidize Cr to Cr^3+. You go through the others in the list and find those that give a + voltage. When you have found the one (or ones) that fit the criteria of oxidizing Cr to Cr^3+, you apply that information with the Ag ==> Ag^+ and find those that will oxidize Cr but not Ag.
b is done exactly the same way but with two different half reactions.
Using a table of standard reduction potentials, determine the best answer to the following questions.
a) Which of the following reagents would oxidize Cr to Cr3 , but not Ag to Ag ?
Choices: Ca^2+, Br2, Co^2+, Ca, Br^-, or Co.
b) Which of the following reagents would oxidize I– to I2, but not Cl– to Cl2?
Choices: Ca^2+, Br2, Co^2+, Ca, Br^-, or Co.
Can someone explain to me how to do this? I don't even know how to start. I got the values but I don't understand how it works.
1 answer