## Predicting if a Metal Will Dissolve in Acid

#### Learning Objective

• Predict whether a metal will dissolve in acid, given its reduction potential

#### Key Points

• Some metals have stronger “replacing” power than others, indicating that they are more likely to reduce.
• Although H2 is not a metal, it can still be “replaced” by some strongly reducing metals.
• The tendency of a metal to “displace” hydrogen gas from acidic solution determines its solubility; if the metal cannot displace hydrogen, it will not be oxidized and will remain insoluble.
• You can determine if a metal will dissolve in acid by comparing the standard reduction potential of the metal to that of hydrogen gas.

#### Terms

• oxidizeTo increase the valence (the positive charge) of an element by removing electrons.
• reduceTo add electrons/hydrogen or to remove oxygen.

## Activity of Metals

Some metals can be considered to be more “active” than others, in the sense that a more active metal can replace a less active one from a solution of its salt. The classic example is of zinc displacing copper:

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

Here, zinc is more active than copper because it can replace copper in solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a coating of finely divided copper. The blue color of the solution diminishes as copper(II) ion is being replaced.

Similar comparisons of other metals have made it possible to arrange them in order of their increasing electron-donating, or reducing, power. This sequence is known as the electromotive, or activity, series of the metals.

• Activity level 1 (highest): Li, K, Ca, Na
• Activity level 2: Mg, Al, Mn, Zn, Fe
• Activity level 3: Ni, Sn, Pb
• Activity level 4 (lowest): Cu, Ag, Pt, Au

## Activity Series

The activity series has long been used to predict the direction of oxidation-reduction reactions. Consider, for example, the oxidation of copper by metallic zinc mentioned above. Zinc is near the top of the activity series, meaning that this metal has a strong tendency to lose electrons. Copper, on the other hand, is a poorer electron donor, and therefore its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the following reaction to proceed in the direction already indicated, rather than in the reverse direction:

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

An old-fashioned way of expressing this is to say that “zinc will replace copper from solution.”

Note that the table also takes the replacement of hydrogen (H2) into account. Although H2 is not a metal, it can still be “replaced” by some strongly reducing metals. The tendency of a metal to “replace” hydrogen gas from acidic solution will determine its solubility in that solution.

## Reduction Potentials

Each half-cell is associated with a potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. In order to express them in a uniform way, we follow the rule that half-cell potentials are always defined for the reduction direction. Therefore, the half-cell potential for the Zn/Zn2+ electrode always refers to the reduction reaction:

$Zn^{2+} + 2e^ - \rightarrow Zn (s)$

In the cell Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s), the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional reduction half-cell potential. These values can be determined using standard reduction potentials, which can often be looked up. Using the standard reduction potentials of a reaction, one can determine how likely a given metal is to accept or donate electrons. For H2, you can quantitatively deduce whether the given metal will dissolve in aqueous solution.

## Example

The net ionic equation for dissolving Zn in HCl would look like this:

$Zn + 2H^+ \rightarrow Zn^{2+} + H_2$

Set up the oxidation and reduction half-reactions with their cell potential:

$Zn \rightarrow Zn^{2+} + 2e^- \ \ \ E^o = 0.76 \ V$

$2H^+ + 2e^- \rightarrow H_2 \ \ \ E^o = 0.00\ V$

Adding the two half-reactions together gives the overall equation and a positive value for E0. This means the reaction is spontaneous and Zn will dissolve in HCl.