Nernst Equation Calculator

Find cell potential or equilibrium membrane potential from concentrations, step by step
E for E0 = 1.10 V, n = 2, [Zn2+] = 0.100 M, [Cu2+] = 1.00 M
Concentration cell with Cu2+ at 0.0100 M and 1.00 M
Nernst potential for K+ with 5.0 mM outside and 140 mM inside at 37 C
Find Q when E = E0 at equilibrium

The Nernst Equation

The Nernst equation corrects a standard electrode potential for concentrations that are not at standard state:

E=EтИШтИТRTnFlnтБбQE = E^{\circ} - \frac{RT}{nF}\ln Q

  • EE тАФ the actual cell potential, in volts.
  • EтИШE^{\circ} тАФ the standard cell potential (all solutes at 1 M, all gases at 1 bar).
  • R=8.314┬аJтАЙmolтИТ1KтИТ1R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}, TT тАФ absolute temperature in kelvin.
  • nn тАФ moles of electrons transferred in the balanced cell reaction.
  • F=96,485┬аCтАЙmolтИТ1F = 96{,}485\ \mathrm{C\,mol^{-1}}, the Faraday constant.
  • QQ тАФ the reaction quotient, products over reactants, each raised to its coefficient. Pure solids and pure liquids are omitted.

At 25 ┬░C the whole prefactor collapses to a single number once the natural log is converted to base 10:

E=EтИШтИТ0.0592nlogтБб10Q(volts,┬аat┬а298.15┬аK)E = E^{\circ} - \frac{0.0592}{n}\log_{10} Q \quad (\text{volts, at } 298.15\ \mathrm{K})

What it assumes. Concentrations stand in for activities, which holds in dilute solution; the temperature is uniform and known; and EтИШE^{\circ} is the value for the same balanced reaction whose nn you used. That 0.0592 belongs to 25 ┬░C only тАФ at 37 ┬░C the base-10 prefactor is 0.0615 V.

Using It for Cells and Membranes

Electrochemical cells

  1. Balance the half-reactions and read off nn, the electrons cancelled.
  2. Write QQ for the overall reaction, omitting solids and pure liquids.
  3. Substitute into E=EтИШтИТ(0.0592/n)logтБб10QE = E^{\circ} - (0.0592/n)\log_{10} Q at 25 ┬░C.
  4. Interpret: E>0E > 0 means the reaction as written is spontaneous. When QQ is small тАФ products scarce тАФ the log is negative and EE rises above EтИШE^{\circ}.

At equilibrium the cell is dead: E=0E = 0 and Q=KQ = K, which gives

logтБб10K=nEтИШ0.0592\log_{10} K = \frac{n E^{\circ}}{0.0592}

Concentration cells

With the same electrode material on both sides, EтИШ=0E^{\circ} = 0 and the potential comes entirely from the concentration difference.

Membrane potentials

For a single ion of charge zz distributed across a membrane, the same algebra gives the equilibrium (Nernst) potential:

Eion=RTzFlnтБб[ion]out[ion]in=61.5┬аmVzlogтБб10[ion]out[ion]inE_{\text{ion}} = \frac{RT}{zF}\ln\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}} = \frac{61.5\ \mathrm{mV}}{z}\log_{10}\frac{[\text{ion}]_{\text{out}}}{[\text{ion}]_{\text{in}}}

at 37 ┬░C. This is the voltage at which that ion's electrical and diffusional driving forces cancel. It describes one ion; a real resting potential mixes several.

Common Mistakes to Avoid

  • Getting nn wrong. nn is the number of electrons transferred in the balanced overall reaction тАФ 2 for Zn+Cu2+\mathrm{Zn + Cu^{2+}}, not 1. It divides the whole correction term.
  • Using 0.0592 away from 25 ┬░C. That constant is RTlnтБб(10)/FRT\ln(10)/F at 298.15 K. At body temperature use 0.0615 V (61.5 mV).
  • Mixing lnтБб\ln and logтБб10\log_{10}. The RT/nFRT/nF form takes lnтБб\ln; the 0.0592 form takes logтБб10\log_{10}. They differ by 2.303.
  • Including solids or the solvent in QQ. Solid zinc and liquid water have unit activity and never appear.
  • Inverting the membrane ratio. EionE_{\text{ion}} uses outside over inside; flipping it flips the sign, turning тИТ89-89 mV into +89+89 mV.
  • Forgetting the ion's charge sign. For ClтИТ\mathrm{Cl^-}, z=тИТ1z = -1, which reverses the result relative to a cation with the same gradient.

Examples

Step 1: Overall reaction: Zn(s)+Cu2+тЖТZn2++Cu(s)\mathrm{Zn}(s) + \mathrm{Cu^{2+}} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Cu}(s), so Q=[Zn2+][Cu2+]Q = \dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} with the solids omitted
Step 2: Q=0.1001.00=0.100Q = \dfrac{0.100}{1.00} = 0.100, so logтБб10Q=тИТ1.000\log_{10} Q = -1.000
Step 3: E=1.10тИТ0.05922(тИТ1.000)=1.10+0.0296E = 1.10 - \dfrac{0.0592}{2}(-1.000) = 1.10 + 0.0296
Step 4: E=1.1296E = 1.1296 V, reported to the 2 decimal places of EтИШE^{\circ}
Step 5: Fewer products than standard state raises the potential, as expected
Answer: E=1.13E = 1.13 V

Step 1: Both electrodes are copper, so EтИШ=0E^{\circ} = 0 V and n=2n = 2
Step 2: The dilute half-cell is the anode; Q=[Cu2+]dilute[Cu2+]concentrated=0.01001.00=0.0100Q = \dfrac{[\mathrm{Cu^{2+}}]_{\text{dilute}}}{[\mathrm{Cu^{2+}}]_{\text{concentrated}}} = \dfrac{0.0100}{1.00} = 0.0100
Step 3: logтБб10(0.0100)=тИТ2.000\log_{10}(0.0100) = -2.000
Step 4: E=0тИТ0.05922(тИТ2.000)=0.0592E = 0 - \dfrac{0.0592}{2}(-2.000) = 0.0592 V
Answer: E=0.0592E = 0.0592 V (59.2 mV)

Step 1: Potassium carries z=+1z = +1, and at 37 ┬░C the prefactor RTlnтБб(10)/FRT\ln(10)/F is 61.5 mV
Step 2: EK=61.51logтБб105.0140E_{\mathrm{K}} = \dfrac{61.5}{1}\log_{10}\dfrac{5.0}{140}
Step 3: 5.0140=0.035714\dfrac{5.0}{140} = 0.035714, and logтБб10(0.035714)=тИТ1.4472\log_{10}(0.035714) = -1.4472
Step 4: EK=61.5├Ч(тИТ1.4472)=тИТ89.0E_{\mathrm{K}} = 61.5 \times (-1.4472) = -89.0 mV
Step 5: The 2 significant figures of 5.0 mM limit the answer
Answer: EK=тИТ89E_{\mathrm{K}} = -89 mV

Frequently Asked Questions

E = E┬░ - (RT/nF) ln Q. It adjusts a standard electrode potential for the actual concentrations in the cell. At 25 ┬░C it simplifies to E = E┬░ - (0.0592/n) log10 Q, with E in volts.

n is the number of moles of electrons transferred in the balanced overall cell reaction тАФ the number that cancels when you add the two half-reactions. For Zn + Cu2+ it is 2. Getting n wrong scales the entire concentration correction.

Q is the reaction quotient of the balanced cell reaction: concentrations (or partial pressures) of products over reactants, each raised to its stoichiometric coefficient. Pure solids and pure liquids are left out because their activity is 1.

At equilibrium the cell can do no more work, so E = 0 and Q equals the equilibrium constant K. Setting E = 0 gives log10 K = nE┬░/0.0592 at 25 ┬░C, which is how standard potentials are turned into equilibrium constants.

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