# What is the voltage of a galvanic cell made with zinc and aluminum at standard conditions?

Aug 22, 2016

${\xi}_{c e l l}^{\circ} = 0.90 V$

#### Explanation:

The standard reduction potentials for zinc and aluminum are listed below:

$Z {n}^{2 +} + 2 {e}^{-} \to Z n \left(s\right) \text{ " " } {\xi}^{\circ} = - 0.76 V$

$A {l}^{3 +} + 3 {e}^{-} \to A l \left(s\right) \text{ " " " } {\xi}^{\circ} = - 1.66 V$

In a galvanic cell, the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment. The second is therefore, forms the cathode compartment.

Since almuninum has the lowest standard reduction potential ${\xi}^{\circ} = - 1.66 V$, therefore, it should be oxidized as follows:

$\textcolor{red}{\text{Oxidation}}$: $A l \left(s\right) \to A {l}^{3 +} + 3 {e}^{-} \text{ " " " } - {\xi}^{\circ} = 1.66 V$

Note that when the reduction equation is reversed, the sign of the standard reduction potential is reversed as well.

Since zinc will be in the cathode compartment, it will get reduced as follows:

$\textcolor{b l u e}{\text{Reduction}}$: $Z {n}^{2 +} + 2 {e}^{-} \to Z n \left(s\right) \text{ " " } {\xi}^{\circ} = - 0.76 V$

The overall reaction in the galvanic cell is the sum of the two half equations; oxidation and reduction:

$\textcolor{red}{\text{Oxidation}}$: $\left(A l \left(s\right) \to A {l}^{3 +} + 3 {e}^{-}\right) \textcolor{p u r p \le}{\times 2} \text{ " " " } - {\xi}^{\circ} = 1.66 V$
$\textcolor{b l u e}{\text{Reduction}}$: $\left(Z {n}^{2 +} + 2 {e}^{-} \to Z n \left(s\right)\right) \textcolor{p u r p \le}{\times 3} \text{ " " } {\xi}^{\circ} = - 0.76 V$
$- - - - - - - - - - - - - - - - - - -$
$\textcolor{g r e e n}{\text{RedOx}}$: $\textcolor{p u r p \le}{2} A l \left(s\right) + \textcolor{p u r p \le}{3} Z {n}^{2 +} \to \textcolor{p u r p \le}{2} A {l}^{3 +} + \textcolor{p u r p \le}{3} Z n \left(s\right) \text{ " } {\xi}_{c e l l}^{\circ} = 0.90 V$

Therefore, the standard cell potential is : ${\xi}_{c e l l}^{\circ} = 0.90 V$

Note that the half equations where multiplied by the corresponding integers (2 and 3) in order to cancel the number of electrons from the overall equation.

Here is a video that further explains this topic:
Electrochemistry | The Standard Reduction Potential.