Wren asked:
help with any of these questions would be greatly appreciated!!
Use the following redox reaction to answer all of the questions:
Sn2+(aq)+Al(s)—>Sn(s)+Al3+(aq)
1. Balance the reaction
2. What is the reduction half in the above reaction?
3. What is the oxidation half?
4. Could the reaction be used to create a galvanic cell as written? If so, what reaction would occur at the anode? What reaction would occur at the cathode?
5. Determine the standard cell potential for the galvanic cell.
July 30th, 2010 at 2:09 pm
Alright, I’ll give it a go.
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1) 1st step, seperate into two half reactions:
Sn2+ –> Sn
Al –> Al3+
2nd, Balance the charges with electrons.
Sn2+ + 2e- –> Sn
(Since there was a 2+ charge on the left side side and no charge on the other, adding two negative charges (or two electrons) to Sn2+, will eliminate the charge to zero, so that it will be equal to the Sn’s zero charge.)
Al –> Al3+ + 3e-
(Same as before, although now there is a postive charge of 3 on the right, and you must make it equal to Al’s zero charge, thus adding three NEGATIVE electrons on the left will eliminate Al3+’s charge)
3rd step: Putting the whole equation together.
Now you have two half-reactions with electrons in them. When you put the two reactions together to make the main equation, you must make the electrons equal.
To do this, we have 3 electrons on one equation ,and two electrons in the other equation. Multiplying the 3 electrons by 2, and 2 electrons by 3, will make both of the electrons of the two reactions 6.
THen you will have qual electrons. Because of this, you must now multiply the whole half reactions with the number too. Thus:
3Sn2+ + 6e- –> 3Sn (multiply whole equation by 3)
2Al –> 2Al3+ + 6e- (Multiply whole equation by 2)
Put them together:
3Sn2+ + 2Al + 6e- –> 3Sn + 2Al3+ + 6e-
Cross out the equal electrons.
3Sn2+ + 2Al –> 3Sn + 2Al3+ Sn.
Sn2+ gains electrons and become Sn, which satisfies the definition of reduction, which is to GAIN electrons.
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3. The oxidation half is Al –> Al3+ + 3e-
Al looses 3 electrons and become Al3+, as you can see from the equation. In loosing electrons, it satisfies the definition of oxidation which is to LOOSE electrons.
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4. Yes, this reaction could be used to create a galvanic cell. Checking the Standard Reduction Table (you must have one), you can see that Tin is above Aluminium.
Tin is a stronger oxidizing agent ( AKA reduction) than aluminium,being that it is higher up in the table, and aluminum is a strong reducing agent (AKA oxidation).
This confirms previous question that Sn2+ reduced, and Al oxidized, and also our balanced equation.
Anodes always oxidize, therefore aluminum would loose electrons at the anode, the reaction being the half reaction of aluminium.
Cathodes always reduce, and thus, Aluminum looses electrons, the reaction being the half reaction of Aluminum.
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5. In determining the standard cell potential, refer to table (you must have this table because you absolutely need it).
( If it were non-standard you would need to use the Nernst equation. But this isn’t it and I’m babbling, please ignore me.)
Looking at Al -> Al3+ + 3e-, it’s there, a down on the chart. Strangely enough, the equation is reversed.
Looking to the Volts column, you will see -1.66 Because this equation is an oxidation equation, not an oxidation equation, you will have to switch the volt’s charge from -1.66 to POSTIVE 1.66. Just flip the sign.
Keep that in mind, and find Sn, which is further up,
There are two Sn equations, choose the one that is exatcly the same as ours. (being not the Sn4+ one)
2e- + Sn2+ -> Sn Volts = -0.14. You have no need to flip anything, since the equation is a reduction equation, same as the one on the chart.
Add the two volts together 1.66 + (-0.14) = 1.52 Volts