Exercise 6K Question 5

[Vivian Leung 1C]

Hi,

I was wondering how we know what the two 1/2 reactions would look like given the reaction in part d of question 5 in Exercise 6K?

For instance, how do we know the reaction goes from

Au+--> Au(s) and Au^3+--> Au(s)

and NOT

Au+--> Au^3+ for one of the reactions instead?

[Sophia Hu 1A]

I think for this problem it is all determined by the Enaught_cell
Au+ + e- --> Au has Enaught = + 1.69 V
Au3+ + 3- --> Au has Enaught = + 1.40 V

The only way to have an overall positive Enaught is through Enaught = 1.69 - 1.40 = 0.29 V.
Therefore, Au+ + e- --> Au is the reduction half-reaction and Au3+ + 3- --> Au must be the oxidation half reaction

I don't think it can be Au+ --> Au3+ because in the appendix there is no reduction half-reaction for this and therefore no calculated voltage. Therefore, it cannot be an option because there is no way to calculate the voltage difference of the cell without an experimental setup. For our problems in 14B, we can only use half-reactions that are in the Appendix because they have a calculated voltage.

I think it also makes sense when setting up the equations of the half-reaction because although it does seem like Au+ is being oxidized to Au3+ when manipulating the reduction half reactions we are able to cancel out the Au(s) so it only ends up on the products and therefore gives us our chemical reaction of interest:
red: 3Au+ + 3e- --> 3Au
ox: Au --> Au3+ + 3-
overall: 3 Au+ (aq) --> 2 Au(s) + Au3+ (aq)

[Vivian Leung 1C]

I think for this problem it is all determined by the Enaught_cell
Au+ + e- --> Au has Enaught = + 1.69 V
Au3+ + 3- --> Au has Enaught = + 1.40 V

The only way to have an overall positive Enaught is through Enaught = 1.69 - 1.40 = 0.29 V.
Therefore, Au+ + e- --> Au is the reduction half-reaction and Au3+ + 3- --> Au must be the oxidation half reaction

I don't think it can be Au+ --> Au3+ because in the appendix there is no reduction half-reaction for this and therefore no calculated voltage. Therefore, it cannot be an option because there is no way to calculate the voltage difference of the cell without an experimental setup. For our problems in 14B, we can only use half-reactions that are in the Appendix because they have a calculated voltage.

I think it also makes sense when setting up the equations of the half-reaction because although it does seem like Au+ is being oxidized to Au3+ when manipulating the reduction half reactions we are able to cancel out the Au(s) so it only ends up on the products and therefore gives us our chemical reaction of interest:
red: 3Au+ + 3e- --> 3Au
ox: Au --> Au3+ + 3-
overall: 3 Au+ (aq) --> 2 Au(s) + Au3+ (aq)

Thank you Sophia!