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Galvanic Cells - Non Standard Conditions

The basic galvanic cell potential equation (\(E^\circ_{cell}=E^\circ_{reduction}−E^\circ_{oxidation}\)) is only valid in standard conditions (25\(^\circ\)C, 1 atm, 1 M). For other situations, we can use the Nernst equation to help us calculate the potential.


  • \(T=\) Temperature in Kelvin
  • \(R=8.314\:J\cdot K^{-1}\cdot mol^{-1}\)
  • \(n=\) no. of electrons exchanged
  • \(F=\) Faraday’s constant
  • \(Q=\frac{[\text{products}]^x}{[\text{reactants}]^y}\)
\[ E=E^\circ−\frac{2.303RT}{nF}\log Q \]

This equation can be simplified further in 25\(^\circ\)C to:

\[ E=E^\circ−\frac{0.0592}{n} \log Q \]

Using the Nernst Equation

There are five steps in using the Nernst equation

  1. Identify the oxidant and reductant
  2. Balance \(e^−\)
  3. Write the overall reaction
  4. Calculate \(E^\circ\)
  5. Calculate \(Q\) being aware of coefficients

Equilibrium and \(K\)

The spontaneous reaction will continue to produce electricity until the reaction is at equilibrium (\(Q=K\))

We can find what our K value is by rearranging out Nernst equation where \(E=0\) and solving for K:

\[ E^\circ=\frac{0.0592}{n} \log K \]


E.g. Calculate \(E^\circ_{cell}\) of a galvanic cell at 25\(^\circ\)C consisting of \(\ce{Ag_{(s)}}\) and 0.0100 M \(\ce{Ag+}\) and \(\ce{Al_{(s)}}\) and 0.25 M \(\ce{Al^{3+}}\)

\(\hskip{1cm} \ce{Ag_{(aq)}+} + \ce{e−} \hskip{0.25cm} -> \ce{Ag_{(s)}}\hskip{0.9cm} +0.80\:V\)

\(\hskip{1cm} \ce{Al_{(aq)}^{3+}} + \ce{3e− -> Al_{(s)}}\hskip{1cm} −1.66\:V\)

\(\hskip{2cm} \text{Reduction} \times3\hskip{1.1cm} \ce{3Ag_{(aq)}+ + 3e− -> 3Ag_{(s)}}\)

\(\hskip{2cm} \text{Oxidation}\hskip{3.8cm} \ce{Al_{(s)} -> Al_{(aq)}^{3+} + 3e−}\)

\(\hskip{2cm} \text{Combined}\hskip{1.9cm} \ce{Al_{(s)} + 3Ag_{(aq)}+ -> Al_{(aq)}^{3+} + 3Ag_{(s)}}\)


\(\hskip{2cm}Q\hskip{3.4cm}\frac{\ce{[Al^{3+}]}}{\ce{[Ag+]^3}} = \frac{2.50}{(0.01)^3} = 2.5\times10^6\)

Putting it all together

\(\hskip{1cm}E=E^\circ−\frac{0.0592}{n}\log Q\)