j4yman
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reduced by copper
The question simply is, which could be reduced by Cu?
Sn2+ (superscript 2+)
Ag+ (superscript +)
Zn2+ (superscript 2+)
How do i figure this out? ...by using some sort of database?
I know copper will spontaneously reduce Ag+
Not sure about the other 2.
The other question 
Consider the following reduction potentials:
Cu2+ + 2e- --> Cu E° = +0.34 V
Pb2+ + 2e- --> Pb E° = -0.13
For a galvanic cell employing the Cu, Cu2+, Pb, Pb2+ couples, calculate the maximum amount of work (in kJ) that would accompany the reaction of one
mole of lead under standard conditions.
________kJ ?
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12AX7
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Ya, look up a table of reduction potentials.
As for the other, that's easy, find the biggest difference (Cu2+ + Pb > Pb2+ + Cu = 0.34 + 0.13 = 0.47V) and calculate the energy two electrons per
atom get. An electron through 1V gains 1eV energy, and there's 6.02 x 10^23 atoms per mole and two electrons exchanged per ion ("2+") and there's
6.25 x 10^21 electrons in one kilocoulomb and C * V = J so you just have to put it together and crank unit conversion to get kJ/mol.
I never liked kJ/mol or any of that crap. eV own.
Tim
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Darkblade48
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Check out a table of reducing potentials, or alternatively, check out a reactivity series table.
I noticed you posted this in another forum (on a different site) as well 
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j4yman
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| Quote: | Originally posted by 12AX7
Ya, look up a table of reduction potentials.
As for the other, that's easy, find the biggest difference (Cu2+ + Pb > Pb2+ + Cu = 0.34 + 0.13 = 0.47V) and calculate the energy two electrons per
atom get. An electron through 1V gains 1eV energy, and there's 6.02 x 10^23 atoms per mole and two electrons exchanged per ion ("2+") and there's
6.25 x 10^21 electrons in one kilocoulomb and C * V = J so you just have to put it together and crank unit conversion to get kJ/mol.
I never liked kJ/mol or any of that crap. eV own.
Tim |
I'm having trouble understanding this. So is V (0.47/2)?
What about C...?
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j4yman
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i got -90.7 kJ
hmmm i used Wmax = -qE
W = -(2 x 96485) x (0.47)
= -90.7 kJ
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