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.

2 Answers
Aug 2, 2018

Answer:

#\approx"1.071 V"#

Explanation:

My work:
Currently, I have calculated via dimensional analysis, and after 5.0 hours there are #\tt{0.56" mol "e^(-)}#. I think I should be calculating for molarity, and adding/subtracting (depending on whether it is reduction or oxidation) the calculated molarity from the original given molarities. After this, I will have the #\sfQ#* in the #\sf{E_(cell)}# formula, and I can solve for that.

Update : I have #0.28# moles of each aqueous ion, which in 1.00 L solution makes for 0.28 M each. The zinc is oxidized, so I add #0.100+0.28=\stackrel{\color(red)(0.380)}{\cancel(0.128)}# M, and the copper, being reduced, becomes #1.90-0.28=1.62# M.

Am I good so far?
*My #Q#, then, would be #\sf{\frac{[Zn^(2+)]}{[Cu^(2+)]}}=\stackrel{\color(red)(0.380)}{\cancel(0.128)}/1.62#


A picture of my work #\sf{\cancel{"(incorrect version)"}}#
webcam

Aug 2, 2018

Seems like #E_(cell) = "1.12 V"#.


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 #"Cu"^(2+)# to produce #"Zn"^(2+)#, increasing #Q_c# over time.

After #"5.0 hr"# on #"3.0 A"# of current, we just need to find how much charge was involved in this forward reaction.

#"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, #"96485 C"# is contained within #"1 mol e"^(-)#, we are looking at:

#54000 cancel"C" xx ("1 mol e"^(-))/(96485 cancel"C")#

#= "0.5597 mols e"^(-)#

Or, since one atom of #"Cu"^(2+)# is consumed over time for every two electrons to reduce it to #"Cu"(s)#,

#0.5597 cancel("mols e"^(-)) xx ("1 mol Cu"^(2+))/(2cancel("mols e"^(-))#

#= "0.280 mols Cu"^(2+)#

were consumed in this #"5.0-hr"# time interval.

Or, since this is in #"1.0-L"# compartments, this means #"0.280 M"# of #"Cu"^(2+)# was consumed. This gives us an idea of how far we have gone towards equilibrium.

#["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# after #"5.0 hr"#, call it #Q_c'#, is now:

#Q_c' = "0.380 M Zn"^(2+)/("1.62 M Cu"^(2+))#

#= ul0.234567#

Now we can find #E_(cell)# after #"5.0 hr"# instead of initially.

#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")#