. For the reaction below at standard conditions, what would be DG° ?2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) ; E° cell = 0.49 V-284000 kJ-284 kJ-142 kJ-142,000 kJ
Question
. For the reaction below at standard conditions, what would be DG° ?2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) ; E° cell = 0.49 V-284000 kJ-284 kJ-142 kJ-142,000 kJ
Solution
To calculate the standard Gibbs free energy change (ΔG°) for the reaction, we can use the Nernst equation:
ΔG° = -nFE°
where:
- n is the number of moles of electrons transferred in the reaction,
- F is the Faraday constant (96485 C/mol), and
- E° is the standard cell potential.
In this reaction, 2Al loses 3 electrons each to form 2Al3+, so n = 2*3 = 6.
Substituting these values into the equation gives:
ΔG° = -(6 mol)(96485 C/mol)(0.49 V) = -283,426 J = -283.426 kJ
So, the closest answer would be -284 kJ.
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