I have Eo Al = +1.66 written as an oxidation.
Eo for H^+ to H2 = 0
Eocell = 1.66 + 0 = 1.66.
Ecell = 1.66 - (0.05916/6)*log Q where
Q = (Al^3+)^2* (pH2)^3/(H^)^6
You didn't show where the 0.00004 came from but shouldn't that be 1.66 v? Second, the 0.0557 is cubed. I went through the calculation and obtained 1.7 v but I have a new calculator and I may have punched the wrong buttons.
I have it set up as
Ecell = 1.66 + (0.05916/6)*log Q and
Q = (6.0461)^2(0.0557)^3/(6.1)^6
2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)
For the above Al-H2 cell, what is the voltage if [Al3+] = 6.0461 M, [H+] = 6.1 M, and the hydrogen pressure is 0.0557 atm. Eo = 1.66 V.
.00004-(.05916/6)*log(7.1e-4)
=.03 WRONG ANSWER I GOT
.00004=(6.0461*6.0461*.0557)/(6.1^6)
Can you please help?
1 answer
0 answers