6L.5 D.

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BeylemZ-1B
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6L.5 D.

Postby BeylemZ-1B » Sat Feb 22, 2020 1:54 pm

“Write the half rxns, the balanced rxn, and the cell diagram for each of the following skeletal equations...
...(d) Au+ —> Au + Au3+

Can someone explain to me how to write the haLf reactions for this ?

DarrenKim_1H
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Re: 6L.5 D.

Postby DarrenKim_1H » Sun Feb 23, 2020 1:11 pm

Reduction (at cathode): Au+(aq) + e- ->Au(s)

Oxidation (at anode): Au+(aq) -> Au 3+(aq) + 2e-

DarrenKim_1H
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Re: 6L.5 D.

Postby DarrenKim_1H » Sun Feb 23, 2020 1:12 pm

So basically, I think you split the Au+ as the half reactant for both parts. From there, writing the half reaction is pretty standard

KnarGeghamyan1B
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Re: 6L.5 D.

Postby KnarGeghamyan1B » Sun Feb 23, 2020 1:15 pm

Just take each of the products, and see where you can place electrons and how many (on reactants side or products) in order to change Au+ into each product. Write each change as a separate equation.

Megan Vu 1J
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Re: 6L.5 D.

Postby Megan Vu 1J » Mon Feb 24, 2020 12:24 pm

For this specific reaction, Au + acts in both the cathode and anode reaction, so you should see both in each reaction.

For cathode, the half reaction is: Au+ + e- -> Au
For anode, the half reaction is" Au3+ + 3e- -> Au+

Then, you should multiply the half reaction of cathode by 3 and flip the anode reaction around.
1.69 V - 1.40 V = 0.29 V.

Alexis Webb 2B
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Joined: Thu Jul 11, 2019 12:15 am

Re: 6L.5 D.

Postby Alexis Webb 2B » Mon Feb 24, 2020 2:45 pm

Megan Vu 1J wrote:For this specific reaction, Au + acts in both the cathode and anode reaction, so you should see both in each reaction.

For cathode, the half reaction is: Au+ + e- -> Au
For anode, the half reaction is" Au3+ + 3e- -> Au+

Then, you should multiply the half reaction of cathode by 3 and flip the anode reaction around.
1.69 V - 1.40 V = 0.29 V.


Where do you find the values for the voltage?

Eileen Si 1G
Posts: 120
Joined: Fri Aug 30, 2019 12:17 am

Re: 6L.5 D.

Postby Eileen Si 1G » Wed Feb 26, 2020 6:59 pm

Alexis Webb 2B wrote:
Megan Vu 1J wrote:For this specific reaction, Au + acts in both the cathode and anode reaction, so you should see both in each reaction.

For cathode, the half reaction is: Au+ + e- -> Au
For anode, the half reaction is" Au3+ + 3e- -> Au+

Then, you should multiply the half reaction of cathode by 3 and flip the anode reaction around.
1.69 V - 1.40 V = 0.29 V.


Where do you find the values for the voltage?


You can find the values for the voltage in Appendix 2B in the back of the textbook (page A16).

ASetlur_1G
Posts: 101
Joined: Fri Aug 09, 2019 12:17 am

Re: 6L.5 D.

Postby ASetlur_1G » Tue Mar 03, 2020 12:31 am

Eileen Si 1G wrote:
Alexis Webb 2B wrote:
Megan Vu 1J wrote:For this specific reaction, Au + acts in both the cathode and anode reaction, so you should see both in each reaction.

For cathode, the half reaction is: Au+ + e- -> Au
For anode, the half reaction is" Au3+ + 3e- -> Au+

Then, you should multiply the half reaction of cathode by 3 and flip the anode reaction around.
1.69 V - 1.40 V = 0.29 V.


Where do you find the values for the voltage?


You can find the values for the voltage in Appendix 2B in the back of the textbook (page A16).


I don't think the voltage for Au3+ + 2e- --> Au+ is given in the back of the book. The solutions manual shows that the oxidation half-reaction is Au --> Au3+ + 3e-


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