1.2 - Standard Electrode Potential

Half-cells and Standard Conditions

• A half-cell consists of an electrode in contact with an electrolyte containing its ions. Electrode potential is the potential difference established at equilibrium between a metal and a solution containing its ions. A single electrode potential cannot be measured directly; it is measured against a reference half-cell.

• Standard electrode potential, $E^0$, is measured relative to the standard hydrogen electrode under:

  • ion concentration $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$;
  • gas pressure $1\mathrm{atm}$ or $101\mathrm{kPa}$;
  • temperature $25{}^{\circ }\mathrm{C}$ or $298\mathrm{K}$;
  • an inert platinum electrode when the half-cell has no conducting solid electrode.

• The standard hydrogen electrode uses platinum, hydrogen gas at $1\mathrm{atm}$ and $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ ${\mathrm{H}}^{+}$ solution. Its value is defined as: $2{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{H}}_2(\mathrm{g})E^0=0.00\mathrm{V}$

• A salt bridge completes the circuit and maintains electrical neutrality through ion movement.

• The hydrogen electrode’s platinum(IV) oxide coating increases its surface area for adsorption of hydrogen and contact with hydrogen ions.

• For zinc, connect a zinc electrode in $1.0\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$ zinc-ion solution to the hydrogen electrode through a salt bridge containing saturated potassium nitrate. The voltmeter reads $0.76\mathrm{V}$; electrons flow from zinc to platinum.

Standard Electrode Potential Series

• Every half-equation in the $E^0$ series is written as a reduction.
• $E^0$ measures a species’ tendency to accept or release electrons:
  • more positive $E^0$ $\to$ greater tendency to accept electrons, undergo reduction and act as a stronger oxidising agent;
  • more negative $E^0$ $\to$ greater tendency of the reduced form to release electrons, undergo oxidation and act as a stronger reducing agent.
• Example: $\mathrm{Z}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Zn}$ has $E^0=-0.76\mathrm{V}$. Against the standard hydrogen electrode, zinc is the negative terminal and is oxidised.
• To predict a spontaneous redox reaction, select the half-cell with the more positive $E^0$ for reduction and reverse the other half-equation for oxidation. Balance electrons before adding the equations.

Standard Reduction Potentials

Reduction half-equation	$E^0$ / V (298 K)
$\mathrm{L}{\mathrm{i}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Li}$	$-3.04$
${\mathrm{K}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{K}$	$-2.92$
$\mathrm{C}{\mathrm{a}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ca}$	$-2.87$
$\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Na}$	$-2.71$
$\mathrm{M}{\mathrm{g}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Mg}$	$-2.38$
$\mathrm{A}{\mathrm{l}}^{3+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Al}$	$-1.66$
$\mathrm{Z}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Zn}$	$-0.76$
$\mathrm{F}{\mathrm{e}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Fe}$	$-0.44$
$\mathrm{N}{\mathrm{i}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ni}$	$-0.25$
$\mathrm{S}{\mathrm{n}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Sn}$	$-0.14$
$\mathrm{P}{\mathrm{b}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Pb}$	$-0.13$
$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{H}}_2$	$0.00$
$\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Cu}$	$+0.34$
${\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-}\rightleftharpoons 4O{\mathrm{H}}^{-}$	$+0.40$
${\mathrm{I}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{I}}^{-}$	$+0.54$
$\mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{F}{\mathrm{e}}^{2+}$	$+0.77$
$\mathrm{A}{\mathrm{g}}^{+}+{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Ag}$	$+0.80$
$\mathrm{B}{\mathrm{r}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2B{\mathrm{r}}^{-}$	$+1.07$
$\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\rightleftharpoons 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$	$+1.33$
$\mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2C{\mathrm{l}}^{-}$	$+1.36$
$\mathrm{Mn}{\mathrm{O}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\rightleftharpoons \mathrm{M}{\mathrm{n}}^{2+}+4{\mathrm{H}}_2\mathrm{O}$	$+1.52$
${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{H}}_2\mathrm{O}$	$+1.77$
${\mathrm{S}}_2{\mathrm{O}}_8^{2-}+2{\mathrm{e}}^{-}\rightleftharpoons 2S{\mathrm{O}}_4^{2-}$	$+2.01$
${\mathrm{F}}_2+2{\mathrm{e}}^{-}\rightleftharpoons 2{\mathrm{F}}^{-}$	$+2.87$

• Copper reacts with silver nitrate: $\mathrm{Cu}+2A{\mathrm{g}}^{+}\to \mathrm{C}{\mathrm{u}}^{2+}+2Ag$, because $E^0(\mathrm{A}{\mathrm{g}}^{+}/Ag)=+0.80\mathrm{V}$ is more positive than $E^0(\mathrm{C}{\mathrm{u}}^{2+}/Cu)=+0.34\mathrm{V}$. Silver ions are the oxidising agent; copper is the reducing agent.
• Copper does not displace magnesium from magnesium nitrate: $E^0(\mathrm{M}{\mathrm{g}}^{2+}/Mg)=-2.38\mathrm{V}$, so magnesium ions are a weaker oxidising agent than copper(II) ions.