solve a galvanic cell

Problem

Given a list of half-reactions of a cell and standard reduction potentials, draw a galvanic cell and calculate its cell potential.

It is an accepted convention to give potential half-reactions as a reduction, so we will be given reduction reactions even though one (or more) will be reversed to make an oxidation reaction.

For instance, consider a half cell with the following half-reactions:

• $\mathrm{Zn}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Zn}(\mathrm{s})$
• $\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}(\mathrm{s})$

We know that one of the reduction half-reactions must be reversed to make a oxidation-reaction pair.

Which one to keep as reduction?

The half-reaction with the largest positive potential will run as written (as a reduction) and other half-reactions will be forced to run in reverse as oxidation reactions. Make sure to negative the potential of oxidation reactions. The cell potential of the cell is simply the sum of the potentials (after necessary negation).

Total number of $\mathrm{e}{}^{-}$ lost must equal the number of electrons gained. Make sure to balance the half-reactions so that the electrons lost by one side & gained by the other are equal.

${\xi }^{\circ }$ does not change when you balance equations

${\xi }^{\circ }$ values aren’t changed when changing coefficients/balancing half reactions–it’s an intensive property–the value doesn’t depend on how many moles there are or how many times a half-reaction occurs.

Example 1

$$\begin{array}{rlrl}2\times (\mathrm{Fe}{}^{3+}+\mathrm{e}{}^{-}& \to \mathrm{Fe}{}^{2+})& & {\xi }^{\circ }=0.77\mathrm{V}\\ -1\times (\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}& \to \mathrm{Cu}(\mathrm{s}))& & {\xi }^{\circ }=0.34\mathrm{V}\\ 2\mathrm{Fe}{}^{3+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{g})& \to 2\mathrm{Fe}{}^{2+}(\mathrm{aq})+\mathrm{Cu}{}^{2+}(\mathrm{aq})& & \end{array}$$ $$\begin{array}{rl}{\xi }_{cell}^{\circ }& ={\xi }_{red}^{\circ }-{\xi }_{ox}^{\circ }\\ & =0.77\mathrm{V}-0.34\mathrm{V}=0.43\mathrm{V}\end{array}$$

Once again, ${\xi }^{\circ }$ values are not changed as an result of changing coefficient.

Example 2

$$\begin{array}{rlrl}-5\times (\mathrm{Fe}{}^{2+}+2\mathrm{e}{}^{-}& \to \mathrm{Fe}(\mathrm{s}))& & {\xi }^{\circ }=-0.44\mathrm{V}\\ 2\times (\mathrm{MnO}{}_4{}^{-}+5\mathrm{e}{}^{-}+8\mathrm{H}{}^{+}& \to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O})& & {\xi }^{\circ }=1.51\mathrm{V}\\ 2\mathrm{MnO}{}_4{}^{-}(\mathrm{aq})+5\mathrm{Fe}(\mathrm{s})+16\mathrm{H}{}^{+}(\mathrm{aq})& \to 2\mathrm{Mn}{}^{2+}(\mathrm{aq})+5\mathrm{Fe}{}^{2+}(\mathrm{aq})+8\mathrm{H}{}_2\mathrm{O}(\mathrm{l})& & \end{array}$$

Sign

A cell will run spontaneously in the direction that produces a positive cell potential

$\mathrm{MnO}{}_4{}^{-}$ will run as the reduction while $\mathrm{Fe}{}^{2+}$ will be the oxidation.

${\xi }_{cell}^{\circ }=1.51\mathrm{V}+0.44\mathrm{V}=1.95\mathrm{V}$