Question

How do you find the E_cell given standard E_cell, Molarity, and the chemical reaction equation?

Find Ecell for an electrochemical cell based on the following reaction
`[MnO_4^-]= 2.50 M`
`[H^+]= 1.50 M`
`[Ag^+]= 0.0090 M`

E∘cell for the reaction is `+0.880V`
`MnO_4^"-"(aq)+4H^+(aq)+3Ag(s)→MnO_2(s)+2H_2O(l)+3Ag^+(aq)`

Answer 1

`1.02268362671V`

Explanation:

Using the Nerst Equation we can can find the `E_"cell"`

The Nernst equation is

`color(blue)(bar(ul(|color(white)(a/a)E_"cell" = E⁰_"cell" - (RT)/(nF)lnQcolor(white)(a/a)|)))" "`

where

`E_"cell"` = cell potential at non-standard state conditions;
`E⁰_"cell"` = cell potential at standard state
`R` = the universal gas constant (`"8.314 J·K"^"-1""mol"^"-1" = "8.314 V·C·K"^"-1""mol"^"-1"`);
`T` = Kelvin temperature;
`F` = Faraday's constant (`"96 485 C/mol e"^"-"`);
`n` = number of moles of electrons transferred in the balanced equation for the cell reaction;
`Q` = reaction quotient for the reaction `"aA + bB ⇌ cC + dD"`

Note: the units of `R` are `"J·K"^"-1""mol"^"-1"` or `"V·C·K"^"-1""mol"^"-1"`.

The moles refer to the “moles of reaction”.

Since we always have 1 mol of reaction, we can write the units of `R` as `"J·K"^"-1"` or `"V·C·K"^"-1"` and ignore the “`"mol"^"-1"`

Calculate the number of moles of electrons transferred in the balanced equation for the cell reaction`n` for this reaction by determining the half reactions

The half reactions are

`3Ag(s) rarr 3Ag^+(aq) + 3e-`
`MNO_4^-)(aq) + 4H^+(aq) + 3e^-) = MnO_2(s) + 2H_2O(l)`

The number of electrons exchanged is 3

`n = 3`

The `E^@` cell for the reaction is 0.880V

Calculate the value for reaction quotient `Q`

`Q = [Ag^+]^3/([MnO4^-][H^+]^4 )`

`Q = (0.009M)^3/([2.5M][1.5M]^4`

`Q = 5.76e-8`

Substitute values into the Nernst equation and solve for `E_"cell"`.

`E_"cell" = E_"cell" = E⁰_"cell" - (RT)/(nF)lnQ = "0.880V" – (8.314 "V"·color(red)(cancel(color(black)("C·K"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(3 color(red)(cancel(color(black)("mol"))) × "96 485" color(red)(cancel(color(black)("C·mol"^"-1")))) × ln( 5.76e-8 )`

`= 0.88V -( -0.14268362671) = 1.02268362671V`