Question

Can someone help with this. I'm confused on how to step this up.
1. If the concentration of fluoride anion and aluminum cation was increased to 5 M, by how much would the measured Ecell change?

𝐹2(𝑔) +2𝑒−→𝐹−(𝑎𝑞).........𝐸o𝑟𝑒𝑑 =2.87
𝐴𝑙^3+ +3𝑒−→𝐴𝑙(𝑠)......𝐸𝑜𝑟𝑒𝑑= −1.66

Answer 1

Each half cell will change by
Ecell = Eocell - [0.0592/n]*log(reduced form concn/oxidized form concn)
For F2 + 2e ==> 2F^- Ered = 2.53 v
Ecell = 2.53 - 0.0592/2* log(1/5^2)

Answer 2

I tried that and I got Ecell = 2.57 but it was wrong.

Answer 3

Instead of me guessing what you've done, spending my time telling you something you've already tried, it's obvious both of us are wasting our time. Show me what the question is and what you've done complete with math. I'll look at this in the morning. Normally I would stay up and help tonight but it's past my bed time and I have a big day tomorrow. I can't mess up tomorrow for lack of rest. Sorry I can't help more tonight.

Answer 4

You can try to write the full equation and apply the following:
Ecell = Eocell - 0.0592/n log Q
n will be 6 . The rxn is 3F2 + 2Al ==> 2Al^3+ + 6F^-
Q = (F^-)^6(Al^3+)^2/(F2)^3(Al)^2
You have 5 M for the F^- and 5 M for Al^3+ as I understand the problem. (Al metal) and (F2 gas) = 1 and Ecell is the 4.53 or what the sum is for Ecell

Answer 5

In addition you obtained 2.57 but that's just for one half cell. The problem wants to know for the CELL, not the half cell so you should calculate the other half cell also, add them together, and see what the DIFFERENCE is. The problem asks for the CHANGE.

Answer 6

Shush Dr. Bob

Answer 7

Stop being rude school is almost over for everyone so we don't need someone being negative and bringing others down. Have a great day and sta safe.