What is the value of E_(cell) at "25"^oC after 5.0 hours of operation at 3.0 A...?
"The std potential (E_(cell)^o ) of a voltaic cell based on the Zn /Cu^(2+) ion reaction
Zn(s)+Cu^(2+)(aq)\toZn^(2+)(aq)+Cu(s)
is "1.10 V" at 25^@ "C" .
What is the value of E_(cell) at "25"^oC after 5.0 hours of operation at 3.0 A, if [Cu^(2+)]_0 is 1.90 M and [Zn^(2+)]_0 is 0.100 M? Assume the volume in both the anode and the cathode compartments is a constant 1.00 L (each)."
See below answer (from me) for my work.
"The std potential (
is
What is the value of
See below answer (from me) for my work.
2 Answers
Explanation:
My work:
Currently, I have calculated via dimensional analysis, and after 5.0 hours there are
Update : I have
Am I good so far?
*My
A picture of my work
webcam
Seems like
Well, the forward reaction was:
"Zn"(s) + "Cu"^(2+)(aq) -> "Zn"^(2+)(aq) + "Cu"(s)
E_(cell)^@ = "0.34 V" - (-"0.76 V")
= +ul"1.10 V"
The reaction is spontaneous in the forward direction, so if one allows it to proceed naturally, it will do so by consuming
After
"3.0 C"/cancel"s" xx (60 cancel"s")/(cancel"1 min") xx (60 cancel"min")/(cancel"1 hr") xx 5.0 cancel"hr"
= "54000 C"
Since according to the Faraday constant,
54000 cancel"C" xx ("1 mol e"^(-))/(96485 cancel"C")
= "0.5597 mols e"^(-)
Or, since one atom of
0.5597 cancel("mols e"^(-)) xx ("1 mol Cu"^(2+))/(2cancel("mols e"^(-))
= "0.280 mols Cu"^(2+)
were consumed in this
Or, since this is in
["Cu"^(2+)]_"currently" = ["Cu"^(2+)]_0 - "0.280 M"
= "1.90 M" - "0.280 M" = "1.62 M Cu"^(2+)
["Zn"^(2+)]_"currently" = ["Zn"^(2+)]_0 + "0.280 M"
= "0.100 M" + "0.280 M" = "0.380 M Zn"^(2+)
The new
Q_c' = "0.380 M Zn"^(2+)/("1.62 M Cu"^(2+))
= ul0.234567
Now we can find
color(blue)(E_(cell)("5.0 hr")) = E_(cell)^@ - (RT)/(nF)lnQ_c'
= "1.10 V" - ("8.314 V"cdotcancel"C"//cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")/(((2 cancel("mol e"^(-)))/cancel"1 mol atoms") cdot 96485 cancel"C"//cancel("mol e"^(-)))ln(0.234567)
= color(blue)("1.12 V")