Science, Tech, Math › Science Nernst Equation Example Problem Calculating Cell Potential in Nonstandard Conditions Share Flipboard Email Print Roland Magnusson / EyeEm / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated November 06, 2019 Standard cell potentials are calculated in standard conditions. The temperature and pressure are at standard temperature and pressure and the concentrations are all 1 M aqueous solutions. In non-standard conditions, the Nernst equation is used to calculate cell potentials. It modifies the standard cell potential to account for temperature and concentrations of the reaction participants. This example problem shows how to use the Nernst equation to calculate a cell potential. Problem Find the cell potential of a galvanic cell based on the following reduction half-reactions at 25 °CCd2+ + 2 e- → Cd E0 = -0.403 VPb2+ + 2 e- → Pb E0 = -0.126 Vwhere [Cd2+] = 0.020 M and [Pb2+] = 0.200 M. Solution The first step is to determine the cell reaction and total cell potential.In order for the cell to be galvanic, E0cell > 0.(Note: Review Galvanic Cell Example Problem for the method to find cell potential of a galvanic cell.)For this reaction to be galvanic, the cadmium reaction must be the oxidation reaction. Cd → Cd2+ + 2 e- E0 = +0.403 VPb2+ + 2 e- → Pb E0 = -0.126 VThe total cell reaction is:Pb2+(aq) + Cd(s) → Cd2+(aq) + Pb(s)and E0cell = 0.403 V + -0.126 V = 0.277 VThe Nernst equation is:Ecell = E0cell - (RT/nF) x lnQwhereEcell is the cell potentialE0cell refers to standard cell potentialR is the gas constant (8.3145 J/mol·K)T is the absolute temperaturen is the number of moles of electrons transferred by the cell's reactionF is Faraday's constant 96485.337 C/mol )Q is the reaction quotient, whereQ = [C]c·[D]d / [A]a·[B]bwhere A, B, C, and D are chemical species; and a, b, c, and d are coefficients in the balanced equation:a A + b B → c C + d DIn this example, the temperature is 25 °C or 300 K and 2 moles of electrons were transferred in the reaction.<br/>RT/nF = (8.3145 J/mol·K)(300 K)/(2)(96485.337 C/mol)RT/nF = 0.013 J/C = 0.013 VThe only thing remaining is to find the reaction quotient, Q.Q = [products]/[reactants](Note: For reaction quotient calculations, pure liquid and pure solid reactants or products are omitted.)Q = [Cd2+]/[Pb2+]Q = 0.020 M / 0.200 MQ = 0.100Combine into the Nernst equation:Ecell = E0cell - (RT/nF) x lnQEcell = 0.277 V - 0.013 V x ln(0.100)Ecell = 0.277 V - 0.013 V x -2.303Ecell = 0.277 V + 0.023 VEcell = 0.300 V Answer The cell potential for the two reactions at 25 °C and [Cd2+] = 0.020 M and [Pb2+] = 0.200 M is 0.300 volts.