The electrochemical series

In a nutshell

An electrochemical series shows you the standard electrode potentials of elements in order of voltage. This allows you to determine element reactivity and reaction feasibility. A reaction is thermodynamically feasible if it has a positive cell potential, however these predictions can sometimes be incorrect.

Elements and standard electrode potentials

Reactive metals will have a more negative standard electrode potential than less reactive metals as they form positive ions more quickly. The reactive metal has reduced the less reactive metal and the less reactive metal has oxidised the reactive metal.

Example

$Li$ has a more negative standard electrode potential than $Fe$ so will form its positive ions more quickly and $Li$ will reduce $Fe$ .

Reactive non-metals will have a more positive standard electrode potential than less reactive non-metals as they form negative ions more quickly. The reactive non-metal has oxidised the less reactive non-metal whereas the less reactive non-metal has reduced the reactive non-metal.

Example

 $F_2$ has a more positive standard electrode potential than $Cl_2$ so will form its negative ions more quickly and $F_2$ will oxidise $Cl_2$ .

An electrochemical series will show you the standard electrode potentials of different species in order of voltage.

Example

half equation	standard electrode potential ( $V$ )
$F_2+2e^-\leftrightharpoons 2F^-$	$+2.87$
$Ag^++e^-\leftrightharpoons Ag$	$+0.80$
$2H^++2e^-\leftrightharpoons H_2$	$0.00$
$Fe^{2+}+2e^-\leftrightharpoons Fe$	$-0.44$
$Li^++e^-\leftrightharpoons Li$	$-3.05$

As the standard electrode potential becomes more negative down the table, the species on the right are more readily oxidised so the species on the left become more stable.

As the standard electrode potential becomes more positive up the table, the species on the left are more readily reduced so the species on the right become more stable.

Predicting the feasibility of reactions

By using the standard electrode potentials of the metal species, you can determine if the reaction is likely to occur in terms of thermodynamics. A thermodynamically feasible reaction should have a positive cell potential.

Procedure

1.	Work out the half equations for the reaction
2.	Decide which species are being oxidised or reduced
3.	Input the standard electrode potentials into the equation
4.	Determine the feasibility of the reaction by the cell potential value

Example

Determine if this reaction is thermodynamically feasible:

$Mg^{2+}+Pb\rightarrow Mg+Pb^{2+}$ 

Half equation	standard electrode potential ( $V$ )
$Mg^{2+}+2e^-\leftrightharpoons Mg$	$-2.36$
$Pb^{2+}+2e^-\leftrightharpoons Pb$	$-0.76$

First, work out the half equations for the reaction. $Mg^{2+}$ needs to gain $2e^-$ to form $Mg$ and $Pb$ needs to lose $2e^-$ to form $Pb^{2+}$ . This gives you the half equations:

$Mg^{2+}+2e^-\rightarrow Mg$

$Pb\rightarrow Pb^{2+}+2e^-$

$Mg$ has been reduced as it has gained electrons and decreased in oxidation number whereas $Pb$ has been oxidised as it has lost electrons and increased in oxidation number.

Now you can input the standard electrode potentials into the following equation:

$E_{cell}=E^\theta _{reduction}-E^\theta_{oxidation}$

 $E_{cell}=-2.36-(-0.76)$

$E_{cell}=-2.36+0.76$

$E_{cell}=-1.60\ V$ 

This reaction is not thermodynamically feasible as it has a negative cell potential of $\underline{-1.60\ V}$ .

Note: The reverse of this reaction would be thermodynamically feasible as it would have a cell potential of $+1.60\ V$ .

Incorrect predictions

When predicting the feasibility of reactions using standard electrode potentials, these can sometimes be incorrect. Standard electrode potentials use standard conditions, therefore if a reaction is carried out under non-standard conditions such as a change in temperature or concentration, this will alter the electrode potentials.

Example

$Co+Ni^{2+}\rightarrow Co^{2+}+Ni$

HALF EQUATION	STANDARD ELECTRODE POTENTIAL ( $V$ )
$Co^{2+}+2e^-\leftrightharpoons Co$	$-0.23$
$Ni^{2+}+2e^-\leftrightharpoons Ni$	$-0.28$

In this reaction, $Co$ has been oxidised and $Ni$ has been reduced. This would produce an $E_{cell}$ value of:

$E_{cell}=E^\theta _{reduction}-E^\theta_{oxidation}$ 

$E_{cell}=-0.28-(-0.23)$

$E_{cell}=-0.28+0.23$ 

$E_{cell}=-0.05\ V$

The cell potential is $\underline{-0.05\ V}$ so is thermodynamically unfeasible. However, if the concentration of $Co$ is increased, the equilibrium will shift to the right and the oxidation potential would become more negative. The cell potential becomes more positive overall so would be thermodynamically feasible.

Alternatively, if the concentration of $Ni^{2+}$ was increased from $1.00\ mol\ dm^{-3}$ , equilibrium will shift to the right and the reduction potential would become less negative, resulting in a more positive cell potential and a thermodynamically feasible reaction.

A reaction may be thermodynamically feasible but not kinetically feasible. A high activation energy or a slow rate of reaction may prevent a reaction occurring. Despite the cell potential showing that the reaction is thermodynamically feasible, if the reaction is not kinetically feasible, it will not occur.