How do I calculate the cell potential?

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How do I calculate the cell potential?

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Answer 1 · 2017-09-26T23:24:46

$E_{c e l l}^{\circ} = E_{r e d}^{\circ} + E_{O x}^{\circ}$ $E_(cell)^@=E_(red)^@+E_(Ox)^@$

Explanation:

For an in-depth explanation see Ernest Z's answer.

To calculate standard cell potential, you use the formula:

$E_{c e l l}^{\circ} = E_{r e d}^{\circ} + E_{o x}^{\circ}$ $E_(cell)^@=E_(red)^@+E_(o x)^@$ ,

where $E_{c e l l}^{\circ}$ $E_(cell)^@$ is the cell potential, $E_{r e d}^{\circ}$ $E_(red)^@$ is the reduction potential and $E_{o x}^{\circ}$ $E_(o x)^@$ is the oxidation potential. These are at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ .

To calculate cell potential in non-standard conditons, the equation is:

$E_{c e l l} = E_{c e l l}^{\circ} - \frac{R T}{n F} ln Q$ $E_(cell)= E_(cell)^@- (RT)/(nF)lnQ$ ,

where

$E_{c e l l}$ $E_(cell)$ = cell potential at non-standard state conditions
$E_{c e l l}^{\circ}$ $E_(cell)^@$ = standard state cell potential
$R$ $R$ = universal gas constant ( $8.314472 J/mol \cdot K$ $"8.314472 J/mol" cdot "K"$ )
$T$ $T$ = absolute temperature (Kelvin scale)
$F$ $F$ = Faraday's constant ( ${96485 C/mol e}^{-}$ $"96485 C/mol e"^-$ )
$n$ $n$ = number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell
Q = reaction quotient for the reaction

$a A + b B \to c C + d D$ $aA + bB -> cC + dD$ ,

so $Q = \frac{{[C]}^{c} {[D]}^{d}}{{[A]}^{a} {[B]}^{b}}$ $Q=([C]^c [D]^d)/([A]^a [B]^b)$

This is the [Nernst equation.](http://www.chem.purdue.edu/gchelp/howtosolveit/Electrochem/Electrochemical_Cell_Potentials.htm)

I hope I helped!

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How do I calculate the cell potential?

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