# 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 $\left[M n {O}_{4}^{-}\right] = 2.50 M$ $\left[{H}^{+}\right] = 1.50 M$ $\left[A {g}^{+}\right] = 0.0090 M$ E∘cell for the reaction is $+ 0.880 V$ MnO_4^"-"(aq)+4H^+(aq)+3Ag(s)→MnO_2(s)+2H_2O(l)+3Ag^+(aq)

Apr 10, 2017

$1.02268362671 V$

#### Explanation:

Using the Nerst Equation we can can find the ${E}_{\text{cell}}$

The Nernst equation is

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

where

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

Note: the units of $R$ are $\text{J·K"^"-1""mol"^"-1}$ or $\text{V·C·K"^"-1""mol"^"-1}$.

The moles refer to the “moles of reaction”.

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

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

$3 A g \left(s\right) \rightarrow 3 A {g}^{+} \left(a q\right) + 3 e -$
MNO_4^-)(aq) + 4H^+(aq) + 3e^-) = MnO_2(s) + 2H_2O(l)

The number of electrons exchanged is 3

$n = 3$

The ${E}^{\circ}$ cell for the reaction is 0.880V

Calculate the value for reaction quotient $Q$

$Q = {\left[A {g}^{+}\right]}^{3} / \left(\left[M n O {4}^{-}\right] {\left[{H}^{+}\right]}^{4}\right)$

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

$Q = 5.76e-8$

Substitute values into the Nernst equation and solve for ${E}_{\text{cell}}$.

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 )

$= 0.88 V - \left(- 0.14268362671\right) = 1.02268362671 V$