b. Use the Nernst equation to calculate the concentration of the unknown solutio
ID: 706736 • Letter: B
Question
b. Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074 V for the galvanic cell with this unknown combined with your copper half-cell. Hint: This problem is similar to practice problem 2 as well as several in your textbook.
Use the unrounded [Cu2+] value of .050179 M and the unrounded value of the constants as listed in the file link near the top of this assignment to avoid round-off error.
Note: You must use the table of reduction potentials given above.
The unknown metal is Zn2+ and I'm supposed to determine it's concentration using the Nerst equation. I have tried plugging all the numbers into the Nerst equation, but I keep getting the wrong answer. Temperature is 23C and the standard reduction potentials for Cu2+ and Zn2+ given are .337V and -.763V respectively. Please help!
b. Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074 V for the galvanic cell with this unknown combined with your copper half-cell. Hint: This problem is similar to practice problem 2 as well as several in your textbook.
Use the unrounded [Cu2+] value of .050179 M and the unrounded value of the constants as listed in the file link near the top of this assignment to avoid round-off error.
Note: You must use the table of reduction potentials given above.
The unknown metal is Zn2+ and I'm supposed to determine it's concentration using the Nerst equation. I have tried plugging all the numbers into the Nerst equation, but I keep getting the wrong answer. Temperature is 23C and the standard reduction potentials for Cu2+ and Zn2+ given are .337V and -.763V respectively. Please help!
Explanation / Answer
Nernest equation,
Zn2+ + 2?e- ---> Zn(s) E0 = -0.7618 v
Cu2+ + 2?e- ----> Cu(s) E0 = 0.337 v
E0cell = E0cathode - E0anode
= (0.337)- (-0.7618)
= 1.1 v
cell reaction: Zn(s) + Cu2+(aq) -----> Zn^2+(aq) + Cu(s)
Ecell = E0cell - (RT/nF)log[[Zn2+]/[Cu2+]]
R = 8.314 j.k-1.mol-1
T = 296 k
F = 96500 c
n = 2(no of electrons transfered)
1.074 = 1.1-((8.314*296)/(2*96500))log(x/0.050179)
x = 5.49
[Zn2+] = 5.49 M
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