Quick Chemistry Question

[bigndog525]

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found it nevermind

 

[bigndog525]

Anyone out there that can help?

not used to calculating conc. from the Nernst. But, anyway here's what I came up with...

I am fairly confused on what to do now. Would E (right) be the .8 volts observed? And which form of the Nernst equation do I use? I feel like I am working with not enough information. Any help is very much appreciated.

 

[wall1two]

what is the exact question?

 

[Teeth12345]

Hmm seems like you might need more info. I think you would need to know at least one of the concentrations to be able to find the Nickel one.

Like Wall said, do you have the exact question available?

 

[bigndog525]

what is the exact question?

Hmm seems like you might need more info. I think you would need to know at least one of the concentrations to be able to find the Nickel one.

Like Wall said, do you have the exact question available?

This was all the information provided and the question is basically worded how I worded it in my first post... it says to estimate the concentration, so I don't think it needs an exact value. Any ideas?

 

[wall1two]

im having a hard time determining what should be cell standard (which seem to be 99% of the problem) the equation is Ecell= Ereduction + E oxidation. left and right wont get to far

my guess is the cell equation is

2Ag + Ni+2 --> 2Ag+1 + Ni so Ag can lose the lone 4s and Ni can fill its d

in that case the .197 volts is the oxidation potential for Ag and the .8v is the reduction potential of Ni.

then can we use a half cell nerst?


.6= .0592/2e- * log (1/Ni2+)

but i dont think using .8 or (.8-.197) for E is correct. im lost

 

[bigndog525]

im having a hard time determining what should be cell standard (which seem to be 99% of the problem) the equation is Ecell= Ereduction + E oxidation. left and right wont get to far

my guess is the cell equation is

2Ag + Ni+2 --> 2Ag+1 + Ni so Ag can lose the lone 4s and Ni can fill its d

in that case the .197 volts is the oxidation potential for Ag and the .8v is the reduction potential of Ni.

then can we use a half cell nerst?


.6= .0592/2e- * log (1/Ni2+)

but i dont think using .8 or (.8-.197) for E is correct. im lost

Idk dude you might be right. How do you solve for that Ni 2+ though and how did you know to put 1 for the red. species?

 

[wall1two]

if the .6 is right, you know that ni has to be 1*10^-12 i think, so the log ( 1/ 1*10^-12) = 12 and .25 *12 gives you 6, but i dont think what i used for E is right.

as for setting up the equation, Ag has a full d and 1 s electron, so it would want to ditch that higher energy electron, making it the reductant and the species that is getting oxidixed so the equation for the whole cell would then be


2Ag(s) + Ni+2(aq) --> 2Ag+1 + Ni(s) so the half cell ion product for Ni :

Q(reduction) = 1/Ni+2 the 1 is for the Ni in solid form which does not appear as a concentration in K or Q expressions