Solving Selective Oxidation with Standard Reduction Potential

  • Thread starter Thread starter salman213
  • Start date Start date
  • Tags Tags
    Oxidation
AI Thread Summary
The discussion centers on determining which cathodic half-reaction allows for the spontaneous oxidation of Sn to Sn2+ without further oxidation to Sn4+. The standard reduction potential for Sn4+ to Sn2+ is +0.15 V, while Sn2+ to Sn is -0.13 V, indicating it is not spontaneous. The participants analyze various half-reactions, concluding that Pb2+ to Pb (+0.13 V) and Fe3+ to Fe (-0.04 V) are suitable choices for the cathode, as they facilitate the desired oxidation of Sn without advancing to Sn4+. The overall consensus emphasizes the importance of comparing standard potentials to ensure the correct reactions are selected. The discussion concludes with a participant seeking confirmation before submitting their assignment.
salman213
Messages
301
Reaction score
1
1. The standard reduction potential for the half-reaction

Sn4+ + 2e- Sn2+
is +0.15 V.
Consider data from the table of standard reduction potentials for common half-reactions, in your text.

For a galvanic cell under standard conditions, which of the following cathodic half reactions would produce, at the anode, a spontaneous oxidation of Sn to Sn2+ but not Sn2+ to Sn4+.

Sn2+ + 2e- Sn
Pb2+ + 2e- Pb
2H2O + 2e- H2 + 2OH-
PbSO4 + 2e- Pb + SO42-
Fe2+ + 2e- Fe
Fe3+ + 3e- Fe



HOW DO I APPROACH THIS PROBLEM?
 
Physics news on Phys.org
any help >
 
Well what are your thoughts on the problem. Also it seems kind of obvious from your choices think about it..
 
Im not really sure how to approach this , i gues i have to look at standard potentials but I am not sure what the question is asking me.
 
Ok I am confused this is wat i did
Sn4+ + 2e- Sn2+
is +0.15 V.

For a galvanic cell under standard conditions, which of the following cathodic half reactions would produce, at the anode, a spontaneous oxidation of Sn to Sn2+ but not Sn2+ to Sn4+.
Sn2+(aq) + 2e− → Sn(s) −0.13

Sn2+ + 2e- Sn -0.13 V
No not spontaeous this is at equilibirum?

Pb2+ + 2e- Pb 0.13 V
Yes because it is spontaneous and the Sn2+ to Sn4+ will occur at the cathode

2H2O + 2e- H2 + 2OH- -0.828 V
No because this will occur at anode since more negative

PbSO4 + 2e- Pb + SO42- -0.359 V
No because this will occur at anode since more negative

Fe2+ + 2e- Fe −0.44 V
No becuase this will occur at anode since more negative

Fe3+ + 3e- Fe -0.04 V
Yes the Sn will be oxidized to Sn2+ since it has lower potential -.13 and it is spontaneous and also the Sn2+ to Sn4+ will not happen at anode since it is much more postive.


I ONLY HAVE 1 MORE TRY TO GET IT RIGHT ON MY ASSIGNMENT ONLINE PLZ HELPPPPPPPPPPPPPPP
 
Last edited:
I think these are right but I don't want to submit them before making sure with someone smart at this forum since i only have one chance left... lol :)

please anyone check!
 
I got it
 
Last edited:
Back
Top