How do you calculate electrochemical cell potential?

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Dec 12, 2016

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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.

#"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"#

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