Question

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)."

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See below answer (from me) for my work.

Answer 1

`\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`

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

Answer 2

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

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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")`