# Question #57f70

May 5, 2015

A Standard Reduction Potential table should have the electrode potentials listed top to bottom from most negative to most positive.

When this is the case, $Z {n}_{\left(s\right)}$ is actually above $C {u}^{2 +}$ in the table, since it has a more negative standard electrode potential.

The more negative an electrode potential is, the stronger a reducing agent will be. Likewise, the more positive an electrode potential is, the stronger an oxidizing agent will be.

$Z {n}_{\left(a q\right)}^{2 +} + 2 {e}^{-} \to Z {n}_{\left(s\right)}$, ${E}^{0} = \text{-0.76 V}$

$C {u}_{\left(a q\right)}^{2 +} + 2 {e}^{-} \to C {u}_{\left(s\right)}$, ${E}^{0} = \text{+0.16 V}$

A more negative electrode potential means is that $Z {n}_{\left(s\right)}$ can reduce $C {u}^{2 +}$, or, In other words, a more positive potential means that $C {u}_{\left(a q\right)}^{2 +}$ can oxidize $Z {n}_{\left(s\right)}$.

Simply put, $C {u}^{2 +}$ is being reduced by $Z {n}_{\left(s\right)}$.

Check out these answers for a more detailed explanation on how standard electrode potentials work: