Avoltaic cell consists of an Al/Al3+ half-cell and a Ag/Ag+ half-cell. Calculate {Ag+} when {Al3+} = 0.795 M and Ecell = 2.411 V. Use reduction potential values of Al3+ = -1.66 V and for Ag+ = +0.80 V

asked by @sarak34 • over 1 year ago • Chemistry • 5 pts

The answer is 0.138 but I can't figure out the steps

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2 answers

We are told that we have a voltaic cell, which is a spontaneous electrochemical cell so that means that the larger voltage belongs to the cathode and the smaller voltage belongs to the anode.

At the anode we have reduction so the equation is:

Al (s) - ---> Al3+ (aq) + 3 e-

At the cathode we have oxidation so the equation is:

Ag+ (aq) + e - ----> Ag (s)

The number of electrons don't match so you multiply the half reaction with the silver by 3 to give you:

3 Ag+ (aq) + 3 e - ---> 3 Ag (s)

When we combine the two reactions and cancel out the intermediate electrons we get:

3 Ag+ (aq) + Al (s) ----> 3 Ag (s) + Al3+ (aq)

Now you say that cell potential under standard conditions: E^o (cell) = Cathode - Anode = 0.80 - (-1.66) = 2.46 V

Now that we have the overall equation and cell potential under standard conditions we use the Nernst equation to find the missing concentration.

Ecell = E^o (cell) - 0.05916/ n [logQ]

Q = products /reactants = [Al3+] /{Ag+]^3

2.411 V = 2.46 V - 0.05916 / 3 e- [logQ]

Use that equation and plug in what you know to solve for the missing concentration of Ag+^3.

Hope this helps! Let us know if you have any other questions.

answered by @sabrina • over 1 year ago


answered by @jaiannam1 • 8 months ago