Asked by Aria
Using the reaction and the E∘ given below
2Co^3+(aq)+2Cl^−(aq)→2Co^2+(aq)+Cl2(g)
E∘=0.46 V
what is the cell potential at 25 ∘C if the concentration are
[Co^3+]= 0.565M ,
[Co^2+]= 0.554M ,
and [Cl^−]= 0.838M
and the pressure of Cl2 is PCl2= 8.50atm ?
Express your answer numerically in volts.
I know I'm suppose to use the equation
E=E∘-(0.0591/n)logQ
but I'm not really sure how I'm suppose to find the moles of electron (n) and Q. Also, I'm not really sure what I should do with the pressure of Cl2.
Thanks in Advance!
2Co^3+(aq)+2Cl^−(aq)→2Co^2+(aq)+Cl2(g)
E∘=0.46 V
what is the cell potential at 25 ∘C if the concentration are
[Co^3+]= 0.565M ,
[Co^2+]= 0.554M ,
and [Cl^−]= 0.838M
and the pressure of Cl2 is PCl2= 8.50atm ?
Express your answer numerically in volts.
I know I'm suppose to use the equation
E=E∘-(0.0591/n)logQ
but I'm not really sure how I'm suppose to find the moles of electron (n) and Q. Also, I'm not really sure what I should do with the pressure of Cl2.
Thanks in Advance!
Answers
Answered by
DrBob222
2Co^3+ goes to 2Co^2+ so that changes by 2e. To check that, 2Cl^- goes to Cl2 and that's a change of 2e. So n = 2.
Q = (Co^2+)^2*pCl2)/(Co^3+)^2(Cl^-)^2
Just plug in the numbers for Q. For pCl2, plug in 8.50 atm.
Q = (Co^2+)^2*pCl2)/(Co^3+)^2(Cl^-)^2
Just plug in the numbers for Q. For pCl2, plug in 8.50 atm.
Answered by
Aria
Thank you so much, it makes a lot more sense now! :)
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