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)#

1 Answer
Apr 10, 2017

Answer:

#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#