# How do you calculate electrochemical cell potential?

Dec 22, 2013

Warning! VERY long answer! You can calculate the cell potential for an electrochemical cell from the half-reactions and the operating conditions.

#### Explanation:

The first step is to determine the cell potential at its standard state — concentrations of 1 mol/L and pressures of 1 atm at 25°C.

The procedure is:

1. Write the oxidation and reduction half-reactions for the cell.

2. Look up the reduction potential, E⁰_"red", for the reduction half-reaction in a table of reduction potentials

3. Look up the reduction potential for the reverse of the oxidation half-reaction and reverse the sign to obtain the oxidation potential. For the oxidation half-reaction, E⁰_text(ox) = "-" E⁰_text(red).

4. Add the two half-cell potentials to get the overall standard cell potential.

E⁰_text(cell) = E⁰_text(red) + E⁰_text(ox)

At the standard state

Let’s use these steps to find the standard cell potential for an electrochemical cell with the following cell reaction.

$\text{Zn(s)" + "Cu"^"2+""(aq)" → "Zn"^"2+""(aq)" + "Cu(s)}$

1. Write the half-reactions for each process.

$\text{Zn(s)" → "Zn"^"2+""(aq)" + "2e"^"-}$
$\text{Cu"^"2+""(aq)" + "2e"^"-" → "Cu(s)}$

2. Look up the standard potential for the reduction half-reaction.

$\text{Cu"^"2+""(aq)" + "2e"^"-" → "Cu(s)"; E⁰_"red} = + 0.339 V$

3. Look up the standard reduction potential for the reverse of the oxidation reaction and change the sign.

$\text{Zn"^"2+""(aq)" + "2e"^"-" → "Zn(s)"; E⁰_text(red) = "-0.762 V}$

$\text{Zn(s)" → "Zn"^"2+""(aq)" + "2e"^"-"; E⁰_"ox" ="+0.762 V}$

4. Add the cell potentials to get the overall standard cell potential.

$\text{Cu"^"2+""(aq)" + "2e"^"-" → "Cu(s)"; color(white)(mmmmmmm)E⁰_"red" = "+0.339 V}$

$\text{Zn(s)" → "Zn"^"2+""(aq)" + "2e"^"-";color(white)(mmmmmmml) E⁰_"ox"=color(white)(l) "+0.762 V}$

$\text{Cu"^"2+""(aq)" + "Zn(s)" → "Cu(s)" + "Zn"^"2+""(aq)"; E⁰_"cell" = "+1.101 V}$

Non-standard state conditions

If the conditions are not standard state (concentrations not 1 mol/L, pressures not 1 atm, temperature not 25°C), we must take a few extra steps.

1. Determine the standard cell potential.

2. Determine the new cell potential resulting from the changed conditions.

a. Determine the reaction quotient, $Q$.
b. Determine $n$, the number of moles electrons transferred in the reaction.
c. Use the Nernst equation to determine ${E}_{\text{cell}}$, the cell potential at the non-standard state conditions.

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 “$\text{mol"^"-1}$ portion of the unit.

Example

Calculate the cell potential for the following reaction when the pressure of the oxygen gas is 2.50 atm, the hydrogen ion concentration is 0.10 mol/L, and the bromide ion concentration is 0.25 mol/L.

${\text{O"_2"(g)" + "4H"^"+""(aq)" + "4Br"^"-""(aq)" → "2H"_2"O(l)" + "2Br}}_{2} \left(l\right)$

1. Write the half-reactions for each process.

$\text{O"_2"(g)" + "4H"^"+""(aq)" + "4e"^"-" → "2H"_2"O(l)}$
$\textcolor{w h i t e}{m m m m m m m l} \text{2Br"^"-""(aq)" → "Br"_2"(l)" + "2e"^"-}$

2. Look up the standard potential for the reduction half-reaction

$\text{O"_2"(g)" + "4H"^"+""(aq)" + "4e"^"-" → 2H_2"O""(l)"; E⁰_"red" = "+1.229 V}$

3. Look up the standard reduction potential for the reverse of the oxidation reaction and change the sign.

$\text{2Br"^"-""(aq)" → "Br"_2"(l)" + "2e"^"-"; E⁰_text(ox) = "-1.077 V}$

4. Add the cell potentials together to get the overall standard cell potential.

$\textcolor{w h i t e}{m m l l} \text{O"_2"(g)" + "4H"^"+""(aq)" + "4e"^"-" → "2H"_2"O(l)"; color(white)(mmmmm)E⁰_text(red) = "+1.229 V}$

color(white)(mmmmmmml)2×["2Br"^"-""(aq)" → "Br"_2"(l)" + "2e"^"-"]; color(white)(mmm)E⁰_text(ox) = "-1.077 V"

$\text{O"_2"(g)" + "4Br"^"-""(aq)" + "4H"^"+""(aq)" → "2Br"_2"(l)" + "2H"_2"O(l)"; E⁰_text(cell) = "+0.152 V}$

5. Determine the new cell potential at the nonstandard conditions.

a. Calculate the value for the reaction quotient, $Q$.

Q = 1/(P_"O₂"["H"^"+"]^4["Br"^"-"]^4) = 1/(2.50 × 0.10^4 × 0.25^4) = 1.0 × 10^6

b. Calculate the number of moles of electrons transferred in the balanced equation.

$n = \text{4 mol electrons}$

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

${E}_{\text{cell" = E_"cell" = E⁰_"cell" - (RT)/(nF)lnQ = "+0.152 V" – (8.314 "V"·color(red)(cancel(color(black)("C·K"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(4 color(red)(cancel(color(black)("mol"))) × "96 485" color(red)(cancel(color(black)("C·mol"^"-1")))) × ln(1.0 × 10^6) = "0.152 V – 0.089 V" = "0.063 V}}$