# Standard reduction potential

• billllib
In summary, the conversation is about standard reduction potentials and understanding half reactions and their corresponding voltages. The main confusion is about the presence of solid and aqueous species in the half reactions, as well as the use of Ksp in redox reactions. The expert suggests that the student may have not fully understood the basics and encourages them to apply their knowledge to solve a specific question involving the equilibrium of a solid AgCl and a solution of chlorides.

#### billllib

http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=jpg::::::/sites/dl/free/0023654666/650262/Standard_Reduction_Potential_19_01.jpg::Standard reduction potentials

The voltage is slightly different in my textbook.

The questions I am trying to answer involve getting both half reactions and adding up the volts. The problem I am not getting the half reactions.

here are the questions s= solid e- = electrons

Cr(s) + 3AgCl(s) CrCl3(aq) + 3Ag(s) <---> One of the half reactions I get and the other I don't.

Cr>Cr3+ + 3e = -.744 (.744 in my text)

Cr3+(aq) + 3e-<>Cr(s) = .744V

Now this is where I have trouble. I think it would be Ag+ (aq) + e- <> Ag(s)

but the problem is the half reaction when I write is this. 3Agcl(s)+e->3AG(s)

Because of the nature of reaction I would eliminate the moles and the cl which gives

Ag+ (s) + e- <> Ag(s).

This is very similar but there solid where there is an "aq" How is this possible

question 2) Cl2+ 2Br> 2Cl- +Br2 two half reactions e- + Cl2<> 2Cl- the closet half reaction in my book is 2e- + Cl2 <> 2Cl- clearly there is a "2e-" Did I assign oxidation numbers wrong. Again I get for the other half reaction I get 2Br(-)<>br2+e- and it should be br2+ 2e<>2Br

Also why do sometimes you remove the moles like ex 1 and ex 2 you keep the moles in the equation?

Can someone answer this by doing the question and posting a picture of the equation and the work so I can understand.

Please tell me if this very messy and I will rewrite it,

Last edited by a moderator:
Yes, it is messy, but at least the first thing is easy to address.

You are in a way right - the reaction is Ag → Ag+. However, in the presence of Cl- Ag+ immediately is removed from the solution, so the concentration of Ag+ is Ksp/[Cl-] (do you see why?). That means standard redox potentials for reactions Ag → Ag+ and Ag + Cl- → AgCl (electrons omitted for brevity) are different.

billllib said:
In class we didn't covered Ksp/[Cl-]

Sure you did, you just don't see it (which sadly means you have not learned the basics well).

Assuming there is a solid AgCl in the equilibrium with the solution of chlorides of concentration [Cl-], what is the concentration of Ag+?

billllib said:
Ksp was not used for or related to redox reactions. Ksp was its own unit.

It is like saying "I don't know how to calculate 3*5+6, multiplication was its own unit". These are not separated things, they all happen at the same time in the solution.

I am almost certain my teacher didn't teach that. I will send you a private pm because I want to keep the class I am in private. So don't link if you find it, The teacher posts all the notes online.

Please respond in a private pm.

I may have accidentally picked the wrong page and it was not assigned. It was assigned sorry.

billllib said:
I am almost certain my teacher didn't teach that.

You were told what Ksp is, that's perfectly enough.

Problem solving doesn't mean "copy the solution you were already shown in the past" but "apply your knowledge to find the solution". And you know everything that is needed here. Even if you were not shown exactly identical problem it is not more difficult than other problems shown in the notes.

I once again suggest that you try to solve this question:

Borek said:
Assuming there is a solid AgCl in the equilibrium with the solution of chlorides of concentration [Cl-], what is the concentration of Ag+?

assuming Ksp=10-10, and [Cl-] = 1 M.