14.11 (e)

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Hannah Guo 3D
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Joined: Fri Sep 29, 2017 7:06 am

14.11 (e)

Postby Hannah Guo 3D » Sun Feb 18, 2018 11:55 am

1. On the left side of the double line (salt bridge), Sn^4+ was reduced to Sn^2+. Does it mean that we can write the reduction reaction on the left side of the salt bridge in cell diagrams?

2. On the right side of the salt bridge, what is the conductor?

Please help! Thank you!

Sohini Halder 1G
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Joined: Thu Jul 13, 2017 3:00 am
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Re: 14.11 (e)

Postby Sohini Halder 1G » Sun Feb 18, 2018 12:57 pm

No, Sn2+ is reduced to Sn4+. You need to flip whatever reduction reaction you get for the anode side to make it an oxidation reaction. You may have gotten confused because Sn4+ is listed before Sn2+ in the cell diagram, but order does not matter there, as long as it is on the left side of the double line you know it is an oxidation reaction

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