# Calculate \sf{E_"cell"} after the cell operates at a temperature of 298 K for 75 minutes at a current of 3.0 A?

## Originally: "The following galvanic cell begins operating with nonstandard concentrations: $\setminus \texttt{M {n}^{2 +}}$ (0.40 M), $\setminus \texttt{C {r}^{3 +}}$ (0.35 M), and $\setminus \texttt{C {r}^{2 +}}$ (0.25 M)... ...Assume both half-cell solutions have a volume of 1.0 L." I have calculated everything up to the consumed moles (and/or molar concentrations, given the volume of 1 liter), but I don't know how to calculate the final concentrations for either of the chromium ions ... (warning: picture is likely to be unclear)

Aug 8, 2018

${E}_{c e l l} = \text{0.67 V}$

Since it's a galvanic cell (voltaic cell), the desired reaction must be spontaneous by definition. So, we need the standard reduction potentials. TWO sources provide these:
WebAssign
Rezofthestory

$\text{Mn"^(2+)(aq) + 2e^(-) -> "Mn} \left(s\right)$, ${E}_{red}^{\circ} = - \text{1.18 V}$
${\text{Cr"^(3+)(aq) + e^(-) -> "Cr}}^{2 +} \left(a q\right)$, ${E}_{red}^{\circ} = - \text{0.50 V}$

My General Chemistry textbook (Tro) also reports $- \text{0.50 V}$ for the chromium(III) half reaction.

To form the spontaneous reaction, we require ${E}_{c e l l}^{\circ} > 0$, so the no-brainer way is that we subtract the less positive (more negative) from the more positive (less negative) ${E}_{red}^{\circ}$.

E_(cell)^@ = -"0.50 V" - (-"1.18 V")

$= + \text{0.68 V}$

This means that we are placing $\text{Mn} \left(s\right)$ at the anode, and the cell notation is going to show electrons flowing from left to right:

overbrace("Mn"(s) | "Mn"^(2+)("0.40 M"))^"Anode" || overbrace("Cr"^(3+) ("0.35 M"), "Cr"^(2+) ("0.25 M") | "Pt"(s))^"Cathode"

(The inert electrode is required for the chromium side since no chromium metal is formed.)

and the reaction is:

${\text{Mn"(s) + 2"Cr"^(3+)(aq) -> 2"Cr"^(2+)(aq) + "Mn}}^{2 +} \left(a q\right)$

The cell will now operate (spontaneously) for $\text{75 min}$ using $\text{3.0 C/s}$, so we will determine how much of each reactant reacted.

75 cancel"min" xx (60 cancel"s")/cancel"1 min" xx "3.0 C"/cancel"s" xx ("1 mol e"^(-))/(96485 cancel"C")

$= {\text{0.1400 mols e}}^{-}$ are involved.

That means

• 0.1400 cancel("mols e"^(-)) xx ("1 mol Mn"^(2+))/(2 cancel("mol e"^(-))) = "0.0700 mols Mn"^(2+) is produced.

• 0.1400 cancel("mols e"^(-)) xx ("1 mol Cr"^(2+))/(1 cancel("mol e"^(-))) = "0.1400 mols Cr"^(2+) is produced.

• 0.1400 cancel("mols e"^(-)) xx ("1 mol Cr"^(3+))/(1 cancel("mol e"^(-))) = "0.1400 mols Cr"^(3+) is consumed.

When the cell starts to operate, it has

• ${\text{0.40 mols"/cancel"L" xx 1.0 cancel"L" = "0.40 mols Mn}}^{2 +}$

• ${\text{0.25 mols"/cancel"L" xx 1.0 cancel"L" = "0.25 mols Cr}}^{2 +}$

• ${\text{0.35 mols"/cancel"L" xx 1.0 cancel"L" = "0.35 mols Cr}}^{3 +}$

So, now it has...

${\text{Mn"(s) + 2"Cr"^(3+)(aq) -> 2"Cr"^(2+)(aq) + "Mn}}^{2 +} \left(a q\right)$

$\text{I"" "-" "" "" "0.35" "" "" "" "0.25" "" "" } 0.40$
$\text{C"" "-" "" "-0.1400" "" "+0.1400" } + 0.0700$
$\text{F"" "-" "" "" "0.21" "" "" "" "0.39" "" "" } 0.47$

At this point we can calculate ${E}_{c e l l}$.

$\textcolor{b l u e}{{E}_{c e l l}} = {E}_{c e l l}^{\circ} - \frac{\text{0.0592 V}}{n} \log Q$

= "0.68 V" - "0.0592 V"/("2 mol e"^(-)//"1 mol Mn"^(2+)) log ((["Cr"^(2+)]^2["Mn"^(2+)])/(["Cr"^(3+)]^2))

$= \text{0.68 V" - "0.0062 V}$

$=$ $\textcolor{b l u e}{\text{0.67 V}}$