# 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.

Where:

• $$T=$$ 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:

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

Example

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 $$\ce{Al^{3+}}$$

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

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

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

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

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

$$\hskip{2cm}E^\circ\hskip{3.2cm}+0.80−(−1.66)=+2.46\:V$$

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

Putting it all together

$$\hskip{1cm}E=E^\circ−\frac{0.0592}{n}\log Q$$
$$\hskip{1cm}E=2.46−\frac{0.0592}{3}\log⁡(2.5\times10^6)$$
$$\hskip{1cm}E=+2.33\:V$$