# How can I calculate galvanic cell potential?

Sep 28, 2014

You calculate galvanic cell potential by subtracting the electrode potentials of each half-cell.

An electrochemical cell (I assume this is what you mean by a galvanic cell) consists of two half - cells. If you know the electrode potentials of each half - cell then you can work out the potential difference ${E}_{c e l l}$ of the cell as a whole.

For example suppose we set up a cell using the two half - reactions:

$Z {n}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s Z {n}_{\left(s\right)}$ ${E}^{0} = - 0.76 V$

$C {u}_{\left(a q\right)}^{2 +} + 2 {e}^{-} r i g h t \le f t h a r p \infty n s C {u}_{\left(s\right)}$ ${E}^{0} = + 0.34 V$

${E}_{c e l l}$ is simply the voltage difference between the standard electrode potentials of the two half - cells.

So

${E}_{c e l l} = \left(+ 0.34 V\right) - \left(- 0.76 V\right) = + 1.1 V$

${E}_{c e l l}$ is an experimentally measured potential difference and is always positive so always subtract the least positive potential from the most positive one.

To get the overall cell reaction you can tell from the electrode potentials that the zinc half - cell will form the negative side of the cell i.e electrons will flow from the zinc half - cell to the copper half - cell.

This gives:

$Z {n}_{\left(s\right)} + C {u}_{\left(a q\right)}^{2 +} \rightarrow Z {n}_{\left(a q\right)}^{2 +} + C {u}_{\left(s\right)}$