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QUESTION

# How do you calculate electrochemical cell potential?

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

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.

"Zn(s)" + "Cu"^"2+""(aq)" → "Zn"^"2+""(aq)" + "Cu(s)"

1. Write the half-reactions for each process.

"Zn(s)" → "Zn"^"2+""(aq)" + "2e"^"-" "Cu"^"2+""(aq)" + "2e"^"-" → "Cu(s)"

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

"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.

"Zn"^"2+""(aq)" + "2e"^"-" → "Zn(s)"; E⁰_text(red) = "-0.762 V"

"Zn(s)" → "Zn"^"2+""(aq)" + "2e"^"-"; E⁰_"ox" ="+0.762 V"

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

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

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

"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_"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_"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" 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.

"O"_2"(g)" + "4H"^"+""(aq)" + "4Br"^"-""(aq)" → "2H"_2"O(l)" + "2Br"_2(l)

1. Write the half-reactions for each process.

"O"_2"(g)" + "4H"^"+""(aq)" + "4e"^"-" → "2H"_2"O(l)" color(white)(mmmmmmml)"2Br"^"-""(aq)" → "Br"_2"(l)" + "2e"^"-"

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

"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.

"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.

color(white)(mmll)"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"

"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 = "4 mol electrons"

c. Substitute values into the Nernst equation and solve for E_"cell".

E_"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"