# What is the value of E_(cell) at "25"^oC after 5.0 hours of operation at 3.0 A...?

## "The std potential (${E}_{c e l l}^{o}$) of a voltaic cell based on the $Z n$/$C {u}^{2 +}$ ion reaction $Z n \left(s\right) + C {u}^{2 +} \left(a q\right) \setminus \to Z {n}^{2 +} \left(a q\right) + C u \left(s\right)$ is $\text{1.10 V}$ at ${25}^{\circ} \text{C}$. What is the value of ${E}_{c e l l}$ at ${\text{25}}^{o} C$ after 5.0 hours of operation at 3.0 A, if ${\left[C {u}^{2 +}\right]}_{0}$ is 1.90 M and ${\left[Z {n}^{2 +}\right]}_{0}$ 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.

Aug 2, 2018

$\setminus \approx \text{1.071 V}$

#### Explanation:

My work:
Currently, I have calculated via dimensional analysis, and after 5.0 hours there are $\setminus \texttt{0.56 \text{ mol } {e}^{-}}$. 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 $\setminus \textsf{Q}$* in the $\setminus \textsf{{E}_{c e l l}}$ 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 $0.100 + 0.28 = \setminus \stackrel{\setminus \textcolor{red}{0.380}}{\setminus \cancel{0.128}}$ M, and the copper, being reduced, becomes $1.90 - 0.28 = 1.62$ M.

Am I good so far?
*My $Q$, then, would be $\setminus \textsf{\setminus \frac{\left[Z {n}^{2 +}\right]}{\left[C {u}^{2 +}\right]}} = \setminus \frac{\stackrel{\setminus \textcolor{red}{0.380}}{\setminus \cancel{0.128}}}{1.62}$

A picture of my work $\setminus \textsf{\setminus \cancel{\text{(incorrect version)}}}$

Aug 2, 2018

Seems like ${E}_{c e l l} = \text{1.12 V}$.

Well, the forward reaction was:

$\text{Zn"(s) + "Cu"^(2+)(aq) -> "Zn"^(2+)(aq) + "Cu} \left(s\right)$

E_(cell)^@ = "0.34 V" - (-"0.76 V")

$= + \underline{\text{1.10 V}}$

The reaction is spontaneous in the forward direction, so if one allows it to proceed naturally, it will do so by consuming ${\text{Cu}}^{2 +}$ to produce ${\text{Zn}}^{2 +}$, increasing ${Q}_{c}$ over time.

After $\text{5.0 hr}$ on $\text{3.0 A}$ of current, we just need to find how much charge was involved in this forward reaction.

$\text{3.0 C"/cancel"s" xx (60 cancel"s")/(cancel"1 min") xx (60 cancel"min")/(cancel"1 hr") xx 5.0 cancel"hr}$

$=$ $\text{54000 C}$

Since according to the Faraday constant, $\text{96485 C}$ is contained within ${\text{1 mol e}}^{-}$, we are looking at:

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

$= {\text{0.5597 mols e}}^{-}$

Or, since one atom of ${\text{Cu}}^{2 +}$ is consumed over time for every two electrons to reduce it to $\text{Cu} \left(s\right)$,

$0.5597 \cancel{{\text{mols e"^(-)) xx ("1 mol Cu"^(2+))/(2cancel("mols e}}^{-}}$

$= {\text{0.280 mols Cu}}^{2 +}$

were consumed in this $\text{5.0-hr}$ time interval.

Or, since this is in $\text{1.0-L}$ compartments, this means $\text{0.280 M}$ of ${\text{Cu}}^{2 +}$ was consumed. This gives us an idea of how far we have gone towards equilibrium.

["Cu"^(2+)]_"currently" = ["Cu"^(2+)]_0 - "0.280 M"

$= {\text{1.90 M" - "0.280 M" = "1.62 M Cu}}^{2 +}$

["Zn"^(2+)]_"currently" = ["Zn"^(2+)]_0 + "0.280 M"

$= {\text{0.100 M" + "0.280 M" = "0.380 M Zn}}^{2 +}$

The new ${Q}_{c}$ after $\text{5.0 hr}$, call it ${Q}_{c} '$, is now:

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

$= \underline{0.234567}$

Now we can find ${E}_{c e l l}$ after $\text{5.0 hr}$ instead of initially.

$\textcolor{b l u e}{{E}_{c e l l} \left(\text{5.0 hr}\right)} = {E}_{c e l l}^{\circ} - \frac{R T}{n F} \ln {Q}_{c} '$

= "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)

$=$ $\textcolor{b l u e}{\text{1.12 V}}$