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14.11 (e)

Posted: Sun Feb 18, 2018 11:55 am
by Hannah Guo 3D
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!

Re: 14.11 (e)

Posted: Sun Feb 18, 2018 12:57 pm
by Sohini Halder 1G
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