(show all your works and make sure that you answer all questions from part a to
ID: 1015079 • Letter: #
Question
(show all your works and make sure that you answer all questions from part a to part e in the picture below)
(T5 pts For the following redox reaction: Au and the overall winne the three balanced equations and their EMF's for each half cell using Appendix D: lbo Will the overall reaction outlined in part (a) be spontaneous or non-spontaneous on what do you base your answer in part (b) Are EMF voltages extensive or intensive properties Bes did you utilize your answer to part (d) when doing the calculations of part (aExplanation / Answer
a) i) Mg2+(aq) + 2 e --> Mg(s) E°r = 2.37 V
ii) Au3+ (aq) + 3 e --> Au(s) E°r = +1.50 V
The half-cell reaction with more positive potential is the reduction half-reaction. In this case, the reduction of gold ions occurs at the cathode. Magnesium is oxidized at the anode.
So, reduction half cell rxn: Au3+ (aq) + 3 e --> Au(s)
oxidation half cell rxn: Mg(s) Mg2+(aq) + 2 e-
To balance the number of electrons in the equations for the two half-cell reactions, multiply the equation for the oxidation reaction by 3 and the equation for the reduction reaction by 2:
3 Mg(s) --> 3 Mg2+(aq) + 6 e
2 Au3+ (aq) + 6 e --> 2 Au(s)
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iii) 3 Mg(s) + 2 Au3+ (aq) --> 3 Mg2+ (aq) + 2 Au(s) ------- Overall cell reaction
E°r (cell) = E°r (cathode) - E°r (anode) = +1.50 V - (-2.37 V) i.e. E°r (cell) = 3.87 V
(b) Since E°r > 0, this cell will react spontaneously
(c) delta G° = -nFE°r (cell) = -ve , since n= moles of electrons transferred in the balanced cell reaction and F= Faraday’s constant.are always positive, and E°r (cell) is also positive, hence the Gibbs free energy change for the cell will be negative, indicating the cell reaction being spontaneous.
(d) Electromotive force (E°r (cell)), measured in volts (V), is equal to the energy per coulomb of charge transferred (V = J/C). So, it is an intensive property.
(e) Because the electromotive force is an intensive property, doubling the cell reaction stoichiometry (or doing any other multiple) will result in the same electromotive force. That is why when calculating the E°r (cell), although we multiply the individual equations by 2 or 3 as per requirement to cancel out the number of electrons, we do not need to change the E°r (cathode) or E°r (anode) accordingly, they stay constant irrespective of the stoichiometric coefficients as done in the calculation (a).
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