Chem 30 Redox question help

[rinny]

Im not the greatest with redox reactions and am having trouble figuring out the net redox equation for the following:

An acidified tin(II) nitrate solution is titrated with potassium permanganate.

Sn(NO3)2 + KMnO4

So I broke it down into following components:

Sn2+ NO3- K+ MnO4- H+

Im thinking:
MnO4 is Strongest Oxidizing Agent
Sn2+ is Strongest Reducing Agent

If this is correct then...well I need help with the rest of the net redox reaction.

Sn2+ + 2e- --> Sn(s)
MnO4- + 8H+ + 5e- --> Mn2+ + 4H20

NET:
Sn2+ + MnO4- + 8H+ +7e- --> Sn(s) + Mn2+ + 4H20

????Is this correct????


Here is a link to a redox table for those wanting to help:
http://www.pembinatrails.ca/shaftesbury/mrdeakin/srp%20table.gif


All help is much appreciated!

[oilerfan4lyfe]

I'll be honest, it's been a long time since I've done this, but I think something is wrong with your equation

In your equation, both Sn and MnO4 are being reduced since the way you're written them they're both gaining electrons. If you look at oxidation numbers, Sn goes from +2 to 0 meaning it gained electrons, and Mn goes from +7 to +2 also meaning it gained electrons.

If MnO4 is the strongest oxidizing agent then it gets reduced meaning the electrons should be on the left side, but if Sn2 is the strongest reducing agent it should be getting oxidized meaning it loses electrons (the electrons should be on the right side of the arrow?)

I'm not 100% sure but that's my best guess...

[Akcard]

It's been far too many years since I've done redox chemistry (even though I'm a chem major)

If I remember correctly, you find your strongest oxidizing agent and strongest reducing agent then electron balance.

You chose the wrong Sn equation from the looks of it since it would be safe to say that Sn(s) will not being precipitating out of your reaction during your titration.

I ended up taking

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O
as the strongest oxidizing reaction

Sn4+ + 2e- --> Sn2+
as the strongest reducing reaction

Since you started with Sn(II) flip the reduction reaction
Sn2+ --> Sn4+ + 2e-

Add the half reactions together after balancing electrons leaves you with

2MnO4- + 16H+ + 5Sn2+ --> 2Mn2+ + 8H2O + 5Sn4+

[Eleanor]

^ listen to this man.

You need to balance your electrons & Sn 2+ is getting oxidized, not reduced.