Question

Calculate the pH of the cathode compartment for the following reaction given Ecell = 3.01 V when [Cr3+] = 0.15 M, [Al3+] = 0.30 M, and [Cr2O72-] = 0.55 M.

2 Al(s) + Cr2O72-(aq) + 14 H+(aq) → 2 Al3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

Answer 1

Ecell = EoCell - 0.05916/n)*log Q

Ecell is given.
Calculate Eocell the usual way
n = 6
Q = (Al^3+)^2(Cr^3+)^2/(Cr2O7^2-)(H^+)^14 and you plug in the concentrations given in the problem to find Q.

Answer 2

how do you find the ph of H+?

Answer 3

i cant get the answer!!

Answer 4

You fill in all of those numbers. (H^+) is the only unknown in the entire equation (I know I said calculate Q but of course you can't do that). After you've solve for H^+ you convert to pH.

The easy way to do that is to start out with Ecell = EoCell - 0.05916/n)*log Q
Then Ecell-Eocell = - (0.05916/n)*log Q
You plug in Ecell and Eocell and n, and get some number = logQ
Solve for Q.
Then set up the equation for Q as shown in the first response and solve for H^+.

Answer 5

how do you solve for H^14?

Answer 6

You have some number = (H^+)^14
You can do it two ways.
The easier way I think is to key in the number, punch the x^y button on your calculator, and key in 0.07142 (that's 1/14), then hit the enter or = button depending upon how you get the answer out of the calculator. Try it on some number you know to see if you have it down. For example, (H^+)^3 = 8 (that's key in 8, hit x^y, key in 0.3333 and =) and that would be (H^+) = 2 OR another example of (H^+)^4 = 81 so (H^+) = 3 (that's key in 81, hit x^y, key in 0.25 (1/4 is 0.25) etc.

The other way is to use logs.
(H^+)^3 = 8
log(H^+)^3 = log 8
3*log(H^+) = log 8
3*log(H^+) = 0.903
log(H^+) = 0.903/3 = 0.301
(H^+) = 10^0.301 = 2

Answer 7

does the ph equal 0?