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Nernst Equation problem

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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.
 

Answers and Replies

  • #2
symbolipoint
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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.
 
  • #3
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Hrm, +3 -> 0. So 3 electrons being transferred, unless you count the coefficient as well.

Would it be 6 electrons transferred?
 
  • #4
Borek
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Calculate each half cell separately, then combine them.
 

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