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

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SUMMARY

The discussion focuses on calculating the cell potential (Ecell) for a magnesium-aluminum electrochemical cell using the Nernst equation. The standard cell potential was calculated as 0.046 V, but the final Ecell was computed as 0.492 V, which was flagged as incorrect by the input program. Participants identified a potential error in the number of electrons transferred, suggesting that three electrons are transferred from aluminum, leading to a total of six electrons when considering stoichiometry. This indicates a need to recalculate Ecell with the correct electron transfer value.

PREREQUISITES
  • Understanding of the Nernst equation and its application in electrochemistry.
  • Familiarity with standard reduction potentials for magnesium and aluminum.
  • Knowledge of calculating reaction quotient (Q) from concentrations.
  • Basic stoichiometry in chemical reactions, particularly in redox processes.
NEXT STEPS
  • Recalculate Ecell using the correct number of electrons transferred for aluminum.
  • Study the Nernst equation in detail, focusing on its application in non-standard conditions.
  • Review the standard reduction potentials for magnesium and aluminum to ensure accurate calculations.
  • Explore electrochemical cell design and factors affecting cell potential.
USEFUL FOR

Chemistry students, electrochemists, and anyone involved in calculating electrochemical cell potentials and understanding redox reactions.

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

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

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