14.13 (d)
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14.13 (d)
Based on the original chemical equation, Au^+ was reduced to Au(s) and was oxidized to Au^(3+), but the answer reverses the anode reaction and yield Au(s)-->Au^(3+) + 3e^-. Can someone please explain why we need to reverse the half-reaction? Thank you!
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Re: 14.13 (d)
The equation must be reversed because the original oxidation reaction is Au3+(aq) + 3e- -> Au (s). This must be reversed into Au (s) -> Au3+(aq) + 3e- so that the Au 3+ ends up in the products side of the cell reaction given in the problem, Au3+ (aq) -> Au(s) + Au3+ (aq).
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Re: 14.13 (d)
I'm stuck on this question too. For the oxidation half-reaction, why is it Au(s)--> Au^3+(aq) + 3e- instead of Au+(aq)--> Au^3+(aq) + 2e-? Since the original reaction makes it seem like Au+(aq) is getting oxidized, not Au(s).
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