Question 13.3: Suppose that we wish to study the possible galvanic corrosio......

First we need the balanced chemical equation, which in this case can be written immediately because two electrons are transferred in each half-reaction (n=2):

$Zn ^{2+}( aq )+ Cr ( s ) \rightarrow Cr ^{2+}( aq )+ Zn ( s )$

Now if we look up the standard reduction potentials, we find

$\begin{array}{ll}Zn ^{2+}( aq )+2\ e ^{-} \rightarrow Zn ( s ) & E^{\circ}=-0.763\ V \\Cr ^{2+}( aq )+2\ e ^{-} \rightarrow Cr ( s ) & E^{\circ}=-0.910\ V\end{array}$

According to Equation 13.2, the cell potential is:

$E_{\text {cell }}^{\circ}=E_{\text {red }}^{\circ}-E_{\text {ox }}^{\circ}$   (13.2)

$E_{\text {cell }}^{\circ}=-0.763\ V -(-0.910\ V )=0.147\ V$

Inserting this in Equation 13.5,

Discussion The most common error students make in this type of problem is forgetting to include n, the number of electrons transferred. Remember that even if you must multiply a chemical equation that represents the reaction given in the standard reduction potential by some value to match the number of electrons between oxidation and reduction, you do not multiply the standard reduction potential. Reduction potentials are not like thermochemical quantities in this regard.

Check Your Understanding If this reaction could be reversed, what would be the value of the free energy change? What does this value tell you about the spontaneity of the reverse process?