Question

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: `\tt{Mn^(2+)}` (0.40 M), `\tt{Cr^(3+)}` (0.35 M), and `\tt{Cr^(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)

Answer 1

`E_(cell) = "0.67 V"`

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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

`"Mn"^(2+)(aq) + 2e^(-) -> "Mn"(s)`, `E_(red)^@ = -"1.18 V"`
`"Cr"^(3+)(aq) + e^(-) -> "Cr"^(2+)(aq)`, `E_(red)^@ = -"0.50 V"`

My General Chemistry textbook (Tro) also reports `-"0.50 V"` for the chromium(III) half reaction.

To form the spontaneous reaction, we require `E_(cell)^@ > 0`, so the no-brainer way is that we subtract the less positive (more negative) from the more positive (less negative) `E_(red)^@`.

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

`= +"0.68 V"`

This means that we are placing `"Mn"(s)` 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:

`"Mn"(s) + 2"Cr"^(3+)(aq) -> 2"Cr"^(2+)(aq) + "Mn"^(2+)(aq)`

The cell will now operate (spontaneously) for `"75 min"` using `"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")`

`= "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

• `"0.40 mols"/cancel"L" xx 1.0 cancel"L" = "0.40 mols Mn"^(2+)`

• `"0.25 mols"/cancel"L" xx 1.0 cancel"L" = "0.25 mols Cr"^(2+)`

• `"0.35 mols"/cancel"L" xx 1.0 cancel"L" = "0.35 mols Cr"^(3+)`

So, now it has...

`"Mn"(s) + 2"Cr"^(3+)(aq) -> 2"Cr"^(2+)(aq) + "Mn"^(2+)(aq)`

`"I"" "-" "" "" "0.35" "" "" "" "0.25" "" "" "0.40`
`"C"" "-" "" "-0.1400" "" "+0.1400" "+0.0700`
`"F"" "-" "" "" "0.21" "" "" "" "0.39" "" "" "0.47`

At this point we can calculate `E_(cell)`.

`color(blue)(E_(cell)) = E_(cell)^@ - "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))`

`= "0.68 V" - "0.0062 V"`

`=` `color(blue)"0.67 V"`