Asked by Ari
So we're doing oxidation-reduction reactions and batteries and voltages.
We did a lab where we would take two metals, put it in a mixture (bleach/water/vinegar/salt) and used a voltmeter to see the DC voltage.
Now we're supposed to 'calculate the potential difference' for some of the combinations, given this info:
"STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT25°C" (google it for the link since jishka won't let me post it)
so I'll just ask for one example: zinc and aluminum. the DC voltage I got was 0.213 if it matters. how do I do it? thanks :)
also if someone knows how to find which copper charge i'm supposed to use since there are three different ones..
We did a lab where we would take two metals, put it in a mixture (bleach/water/vinegar/salt) and used a voltmeter to see the DC voltage.
Now we're supposed to 'calculate the potential difference' for some of the combinations, given this info:
"STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT25°C" (google it for the link since jishka won't let me post it)
so I'll just ask for one example: zinc and aluminum. the DC voltage I got was 0.213 if it matters. how do I do it? thanks :)
also if someone knows how to find which copper charge i'm supposed to use since there are three different ones..
Answers
Answered by
DrBob222
First, the usual one to use for Cu is
Cu ==> Cu^2+ + 2e
The potential for the Al/Zn couple, using standard reduction potential (but from my text which is several years old--it may not match your data).
Al + 3e = Al(aq) Eo = about -1.66 red pot
Zn + 2e ==> Zn(aq) Eo = -0.763
Reverse the more negative value and add to the other one; ie.,
Al ==> Al^3+ + 3e 1.66
Zn^2+ + 2e ==> Zn -0.763
Sum = 1.66 + (-0.763) = 0.897 or close to that is what you should have obtained in the lab IF you used the right measuring equipment.
Cu ==> Cu^2+ + 2e
The potential for the Al/Zn couple, using standard reduction potential (but from my text which is several years old--it may not match your data).
Al + 3e = Al(aq) Eo = about -1.66 red pot
Zn + 2e ==> Zn(aq) Eo = -0.763
Reverse the more negative value and add to the other one; ie.,
Al ==> Al^3+ + 3e 1.66
Zn^2+ + 2e ==> Zn -0.763
Sum = 1.66 + (-0.763) = 0.897 or close to that is what you should have obtained in the lab IF you used the right measuring equipment.
Answered by
DrBob222
And if the solutions were 1M.
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