a) Use the Nernst equation to calculate the Ni/Ni^+2 half cell potential. Do the same for the Al^+3/Al half cell potential.
Add the two to obtain Ecell. I get -1.45v if I didn't make an error. You need to confirm that. Also you need to write the balanced equation for the cell AS WRITTEN, keeping in mind that it is negative which means it will not proceed spontaneously as written and as drawn.
For part b, the directions tell you to make Al the oxidized electrode which means the cell is rearranged and will proceed spontaneously; you will need to adjust the signs of the numbers (as well as the equation) to fit.
Ecell = 1.54v
n = 6 electrons.
You can fill in the values for components in Q.
Ecell = Eocell -(0.0592/n)log Q
An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode.
(a) What is the potential of this cell at 25°C if the aluminum electrode is placed in a solution in which [Al3+] = 7.3 10-3 M?
(b) When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25°C is 1.54 V. Calculate [Al3+] in the unknown solution. (Assume Al is oxidized.)
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