Can Fe2+(aq) Reduce Sn2+(aq) to Sn Metal?

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Fe2+(aq) cannot reduce Sn2+(aq) to Sn metal due to the incorrect application of standard reduction potentials. The standard reduction potential for Sn2+ is -0.14 V, while for Fe2+ it is -0.44 V. To act as a reducing agent, Fe2+ must be oxidized to Fe3+, which means the correct half-reaction must be used. The overall cell potential calculation must reflect the oxidation of Fe2+ and the reduction of Sn2+, leading to a balanced reaction equation.

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Is Fe2+(aq) capable of reducing Sn2+(aq) to Sn metal?



The standard reduction potentials from my book are -0.14 for Sn2+ and -0.44 for Fe2+



Because Sn has a more positive Epot it has a greater tendency to be reduced. So add the Epot up: -0.14 + 0.44 (oxidation)= 0.3. It's positive so this reaction can occur. However my answer key says the opposite. Can someone explain please? Thanks.
 
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You are looking at the wrong potential for iron. -0.44 means iron gets reduced to metallic Fe, you want it to be a reducing agent - that means it has to be oxidized to Fe3+.

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Borek wrote: You are looking at the wrong potential for iron. -0.44 means iron gets reduced to metallic Fe, you want it to be a reducing agent - that means it has to be oxidized to Fe3+.

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Yes I know. Here's the way I have it set up:
Fe --> Fe2+ + 2e Eox = 0.44V
Sn2+ + 2e --> Sn Ered = -0.14V

Adding the half reactions I get Ecell = 0.44-0.14 = 0.3 which is positive so the reaction goes. Apparently this is wrong though. I'm confused...
 
You know you are wrong, yet you repeat the same mistake? You have selected wrong reaction for iron. If you have a solution containing Fe2+ and Sn2+ you can't get both reduced to metal at the same time. If one gets reduced, the other has to be oxidized.

Try to write balanced reaction equation. Remember Fe2+ and Sn2+ are your rectants.

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methods
 
Oh I see! Thanks so much Borek!
 

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