**Practice Problem 10**

Calculate the equilibrium constant at 25^{o}C for the
reaction between zinc metal and acid.

Zn(

s) + 2 H^{+}(aq) Zn^{2+}(aq) + H_{2}(g)

**Solution**

We start by calculating the overall standard-state cell potential for this reaction.

Zn Zn^{2+} + ^{-} |
E^{o}_{ox} = -(-0.7628
V) |
||

+ 2 H^{+} + ^{-}
H_{2} |
E^{o}_{red} = 0.0000 V |
||

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Zn(s) + 2 H^{+}(aq)
Zn^{2+}(aq) + H_{2}(g) |
E^{o} = 0.7628 V |

We then substitute into the Nernst equation the implications
of the fact that the reaction is at equilibrium: *Q*_{c}
= *K*_{c} and *E* = 0.

We then rearrange this equation

and solve for the natural logarithm of the equilibrium constant.

Substituting what we know about the reaction into this equation gives the following result.

This is a very large equilibrium constant, which means that equilibrium lies heavily on the side of the products. The equilibrium constant is so large that the equation for this reaction is written as if it proceeds to completion.

Zn(*s*) + 2 H^{+}(*aq*)
Zn^{2+}(*aq*)
+ H_{2}(*g*)