Calculate Ecell using Nernst Equation and Appendix D Data | Mg-Al Cell Problem

[User Name]

Homework Statement

Use the Nernst equation and data from Appendix D in the textbook to calculate E cell for each of the following cells.

Mg(s)|Mg+2 (0.012 M) || [Al(OH)4]- (0.25M), OH- (0.048M)|Al(s)

Homework Equations

Ecell = Ecell(standard) - 0.0591/n * log Q

The Attempt at a Solution

Half reaction of Mg: -2.356
Half reaction of [Al(OH)4]-: -2.310

Ecell (standard) = -2.310-(-2.356) = .046 V

Equation written out in spontaneous form would be...

3Mg + 2[Al(OH)4]- -> 3Mg+2 + 8OH- + 2Al

Then solving for Q... which is [product]/[reactant] would be...

Q = ([.048]^8 * [.012]^3)/[.25]^2 = 7.79E-16

Number of moles of electrons transferred is 2.

Plug everything in...

Ecell = .046 - (.0591/2)*log (7.79E-16)

Which gives me

Ecell = 0.492 V

However, the program I'm entering this into says it's incorrect.

Any pointers on where I went wrong?

Thanks in advance.

 

[symbolipoint]

Mg(s)|Mg+2 (0.012 M) || [Al(OH)4]- (0.25M), OH- (0.048M)|Al(s)
...
However, the program I'm entering this into says it's incorrect.

Any pointers on where I went wrong?

3Mg + 2[Al(OH)4]- -> 3Mg+2 + 8OH- + 2Al

Then solving for Q... which is [product]/[reactant] would be...

Q = ([.048]^8 * [.012]^3)/[.25]^2 = 7.79E-16

Number of moles of electrons transferred is 2.

Recheck how many electrons transfer for the Aluminim.

 

[User Name]

Hrm, +3 -> 0. So 3 electrons being transferred, unless you count the coefficient as well.

Would it be 6 electrons transferred?

 

[Borek]

Calculate each half cell separately, then combine them.