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Topic: Voltaic cell  (Read 2798 times)

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Offline chrissy

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Voltaic cell
« on: October 09, 2009, 01:26:47 AM »
Here is another problem i couldn't get the right answer for but don't know the actual answer :

Here is what you have at your disposal to make a voltaic cell: a voltmeter, a power supply, two loarge beackers, a U-shaped tube with two porous plugs, wires, 1mol / L aqueaous solution of tin (II) chloride, copper(II chloride and sodium chloride, as well as tin, copper and platinum electrodes. The standard reduction potentials of tin(II) and copper(II) ions are - 0.13V and 0.15V, respectively.

a) Write balanced equations for the two half reactions and the overall reaction and find the voltaic cell's standard potential

what i got :

oxidation: Sn(s) --> Sn+2(aq) + 2e              Standard potential (E*degree) : 0.13 (because i know you're supposed to flip the lowest # of the two)
reduction: Cu+2 (aq) + 2e --> Cu(s)                             Standard potential : 0.15
overall: Sn(s) + Cu+2(aq)+2e --> Sb+2 + Cu(s) +2e         Standard potential for the cell : 0.2 ( here i dont know if i have to subtract or add)

b) which half-reaction would occur at the cathode?

Cu+2(aq)+2e --> Cu(s)

c) You have electrodes made of three different metals. Which one(s) would be appropriate to use as the cathode in your voltaic cell?

Tin? Why?
Copper? Why?
Platinum? Why?


Thanks a bunchhh :)

Offline Dan

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Re: Voltaic cell
« Reply #1 on: October 09, 2009, 08:11:15 AM »
Ok, my electrochem is extremely rusty but I'll try to give you a hand... Someone step in if I'm making a fool of myself...

a) Write balanced equations for the two half reactions and the overall reaction and find the voltaic cell's standard potential

what i got :

oxidation: Sn(s) --> Sn+2(aq) + 2e              Standard potential (E*degree) : 0.13 (because i know you're supposed to flip the lowest # of the two)
reduction: Cu+2 (aq) + 2e --> Cu(s)                             Standard potential : 0.15
overall: Sn(s) + Cu+2(aq)+2e --> Sb+2 + Cu(s) +2e         Standard potential for the cell : 0.2 ( here i dont know if i have to subtract or add)

How did you get 0.2?

There are two ways you can look at the calculation, bearing in mind that the reaction is a combination of an oxidation and a reduction (redox):

1. Reduction + Oxidation
2. Reduction(of the less reducing metal) - Reduction(of the more reducing metal)

Note that Oxidation = - Reduction

Quote
b) which half-reaction would occur at the cathode?

Cu+2(aq)+2e --> Cu(s)

Yup, you get reduction at the cathode so I agree with your answer here.

Quote
c) You have electrodes made of three different metals. Which one(s) would be appropriate to use as the cathode in your voltaic cell?

Think about what you are producing at the cathode, and the solution it is in contact with... Are there any of those metals you would definitely not use?
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