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The EMF of a cell

  • #1
Krushnaraj Pandya
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71

Homework Statement


The EMF of the cell : Ni(s) | Ni2+ (0.1M) || Au 3+ (1.0M) | Au (s) is

[ given E°(Ni2+/Ni)= -0.25V; E° (Au3+ / Au )= +1.50V ]

Homework Equations


E°cell=E°(cathode)-E°(anode) For SRP.
Nernst equation

The Attempt at a Solution


SOP is given, therefore converting to SRP we take minus sign of both. Now Ni to Ni2+ is at anode and Au3+ to Au is at cathode. Therefore E(cell)= -1.5-0.25= -1.75V
Writing balanced reaction 3Ni+2Au3+ gives 3Ni2+ + 2Au, no. of electrons transferred=6.
Putting into nernst equation, EMF should be -1.75-(0.059/6)log(0.1)^3 which gives the wrong answer. Where am I wrong, I'd appreciate some help
 

Answers and Replies

  • #2
mjc123
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Why have you got log(0.1)^3 ? Write out the Nernst equation in full for this cell.
 
  • #3
Krushnaraj Pandya
Gold Member
697
71
Ok, so we have E°=-1.75, RT/F=0.059, Q=[Ni2+]^3/[Au3+]^2=(0.1)^3/(1)^2=(0.1)^3
replacing in E=E°-2.303(RT/nF)logQ we get the equation I wrote in"Attempt at a solution"
 
  • #4
mjc123
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Sorry, I misread the question, I thought [Au3+] was also 0.1M.
I think you've got the signs wrong. E for the cell is ER - EL by convention
Ecell = E(Au3+/Au) - E(Ni2+/Ni)
= E0(Au3+/Au) - E0(Ni2+/Ni) + RT/nF*log([Au3+]2/[Ni2+]3)
 
  • #5
Krushnaraj Pandya
Gold Member
697
71
Sorry, I misread the question, I thought [Au3+] was also 0.1M.
I think you've got the signs wrong. E for the cell is ER - EL by convention
Ecell = E(Au3+/Au) - E(Ni2+/Ni)
= E0(Au3+/Au) - E0(Ni2+/Ni) + RT/nF*log([Au3+]2/[Ni2+]3)
The magnitude should remain the same (note that reversing the terms inside log reverses its sign)
 
  • #6
Krushnaraj Pandya
Gold Member
697
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Alright, I realized my mistake. I got confused in sign conventions. Thank you for your help :D
 
  • #7
mjc123
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You might find it helpful
(i) to do the Nernst equation on each electrode separately; that way you ought to get the magnitude of the cell potential right, even if you get the sign convention wrong.
(ii) to think Le Chatelier - if I increase the concentration of X, will that pull the equilibrium to the right or left - increase or decrease cell potential?
 
  • #8
Krushnaraj Pandya
Gold Member
697
71
You might find it helpful
(i) to do the Nernst equation on each electrode separately; that way you ought to get the magnitude of the cell potential right, even if you get the sign convention wrong.
(ii) to think Le Chatelier - if I increase the concentration of X, will that pull the equilibrium to the right or left - increase or decrease cell potential?
Very helpful tips, thank you
 

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