Calculate Delta G for the spontaneous cell produced from the following:
Al^3+ + 3e^- -> Al (s)
Mg^2+ + 2e^- -> Mg (s)
In order to solve this, you need to determine the half reactions..ie which is reduced and which is oxidized. How do I know which one is being reduced and which is being oxidized? :s
deltaG= -nFEcell
n= number of electrons transferred in the reaction
3Mg= 3Mg+2 + 6e-
2Al+3 + 6e- = 2Al
n= 6
F= faradays constant= 96500 J/V mol e-
Ecell= Ecathode-Eanode
Cathode: Al+3 + 3e- = Al E= -1.66 V
Anode: Mg+2 + 2e- = Mg E= -2.37
Ecell= -1.66 - -2.37
Ecell= .71
deltaG= - (6) (.71) (96500)
deltaG= -410000 J/mol or -410 kJ/mol
You look up the Eo values for Al and Mg. Some quick numbers I remember, but you need to look them up and do the work more accurately,is
Eo for Al^+3 + 3e = -1.66 volt
Eo for Mg^+2 + 2e = -2.37
Turn the most negative value around to make it an oxidation and add to the other (in order to obtain a positive value for the cell potential).
3(Mg ==> Mg^+2e) E = +2.37
2(Al^+3 + 3e ==> Al) E = -1.66
=====================
3Mg + 2Al^+3 + 6e ==>3Mg^+2 + 2Al + 6e
Then, remember oxidation is the loss of electrons; reduction is the gain of electrons.
To determine which species is being reduced and which one is being oxidized, you can follow a few steps:
Step 1: Write down the oxidation states for each element in the reaction. In this case, we have:
Al^3+: +3 oxidation state
Al (s): 0 oxidation state
Mg^2+: +2 oxidation state
Mg (s): 0 oxidation state
Step 2: Compare the oxidation states of the elements before and after the reaction. The element that undergoes an increase in oxidation state is being oxidized, while the element that undergoes a decrease in oxidation state is being reduced.
In the given reactions:
- The oxidation state of aluminum (Al) changes from +3 to 0. It goes from a positive oxidation state to no charge, so it is being reduced (gaining electrons).
- The oxidation state of magnesium (Mg) changes from +2 to 0. It goes from a positive oxidation state to no charge, so it is also being reduced (gaining electrons).
Both aluminum and magnesium are being reduced in their respective half-reactions.
Remember that the species being oxidized loses electrons while the species being reduced gains electrons.
To determine which species is being oxidized and which is being reduced, you need to look at the changes in oxidation states of the atoms involved.
In the given half-reactions:
Al^3+ + 3e^- -> Al (s)
Mg^2+ + 2e^- -> Mg (s)
The oxidation state of Al changes from +3 to 0, and the oxidation state of Mg changes from +2 to 0.
Understanding oxidation states:
- Oxidation state refers to the charge an atom would have in a compound if electrons were completely transferred.
- Atoms tend to gain or lose electrons in a way that allows them to attain a noble gas electron configuration.
In the first half-reaction, Al^3+ is reduced (gains electrons) since the oxidation state of Al decreases from +3 to 0.
In the second half-reaction, Mg^2+ is also reduced (gains electrons) since the oxidation state of Mg decreases from +2 to 0.
Therefore, Al^3+ is being reduced, and Mg^2+ is also being reduced.