HW 14.13d

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HW 14.13d

Postby chemstudent_1K » Fri Feb 05, 2016 12:26 am

Part of HW question 14.13d asks to write the half reactions for Au+->Au + Au+3

One of the half reactions is Au+ + e- -> Au, but how do we know to use Au+3 + 3e- -> Au for the other half reaction instead of Au+3 + 2e- -> Au+?

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Re: HW 14.13d

Postby Chem_Mod » Sat Feb 06, 2016 10:19 am

The charge has to be balanced and the way you suggested will result in the overall reaction to be 3Au+ --> Au3+ + 2 Au

Will Tsai 3C
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Re: HW 14.13d

Postby Will Tsai 3C » Sat Feb 06, 2016 5:48 pm

I have the same question too. If Au(s) is not included on the product side of the original equation, how come we can assume that it will be in the half reaction for the anode? (Au3+(aq) + 3e- -> Au(s)

Sohini Halder 1G
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Re: HW 14.13d

Postby Sohini Halder 1G » Sun Feb 18, 2018 12:54 pm

I still don't understand this setup. Do we have to assume that Au(s) is present in both the reduction and oxidation reactions because it is the conductive metal electrode in the solution? And then knowing this, is this why Au(s) goes to both Au 1+ and Au 3+?

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