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)#
Find Ecell for an electrochemical cell based on the following reaction
E∘cell for the reaction is
1 Answer
Explanation:
Using the Nerst Equation we can can find the
The Nernst equation is
#color(blue)(bar(ul(|color(white)(a/a)E_"cell" = E⁰_"cell" - (RT)/(nF)lnQcolor(white)(a/a)|)))" "#
where
Note: the units of
The moles refer to the “moles of reaction”.
Since we always have 1 mol of reaction, we can write the units of
Calculate the number of moles of electrons transferred in the balanced equation for the cell reaction
The half reactions are
The number of electrons exchanged is 3
The
Calculate the value for reaction quotient
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 ) #