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

\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)"}}
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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")