Standard Electrode Potential

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Standard Electrode Potential

We can place redox systems in order of their oxidising/reducing ability if we measure their electrode potential against the standard hydrogen electrode, which has a potential of 0.00V.

Note: Standard conditions must be adhered to, so for example if using copper rods they must be immersed into a solution of 1 mol dm-3 copper ions.

By convention, we define the e.m.f of an electrochemical cell as follows:

Eθ = Eθ right-hand half-cell - Eθ left-hand half-cell.

The s.h.e is always placed on the left-hand side.

The standard electrode potential of a metal is the potantial acquired when the metal is immersed in a 1 mol dm-3 solution of its ions at a temperature of 25oC.

It has the symbol Eθ and the units V.

As stated previously, we can tabulate the order of oxidising/reducing ability of a system, this we call the electrochemical series.

The most positive E value is at the top, i.e. the greatest oxidising agent. The most negative value E at the bottom i.e. the greatest reducing agent.

Reaction   Eo (298K)/V
F2(g) +2e- &ruldhar; 2F-(aq) +2.87
Co3+(aq) + e- &ruldhar; Co2+(aq) +1.82
Pb4+(aq) +2e- &ruldhar; Pb2+(aq) +1.69
Mn3+(aq) + e- &ruldhar; Mn2+(aq) +1.51
Au3+(aq) +3e- &ruldhar; Au(s) +1.42
Cl2(aq) +2e- &ruldhar; 2Cl-(aq) +1.36
Br2(aq) +2e- &ruldhar; 2Br-(aq) +1.07
NO3-(aq) + 2H+(aq) + e- &ruldhar; NO2(aq) + H2O(I) +0.81
Ag+(aq) + e- &ruldhar; Ag(s) +0.80
Fe3+(aq) + e- &ruldhar; Fe2+(aq) +0.77
I2(aq) +2e- &ruldhar; 2I-(aq) +0.54
Cu+(aq) + e- &ruldhar; Cu(s) +0.52
02(g) + 2H2O(l) +4e- &ruldhar; 4OH-(aq) +0.40
Cu2+(aq) +2e- &ruldhar; Cu(s) +0.34
S042-(aq) + 4H+(aq) +2e- &ruldhar; SO2(aq) + 2H20(l) +0.17
Cu2+(aq) + e- &ruldhar; Cu+(aq) +0.15
Sn4+(aq) +2e- &ruldhar; Sn2+(aq) +0.15
Fe3+(aq) +3e- &ruldhar; Fe(s) +0.04
2H+(aq) +2e- &ruldhar; H2(g) 0.00
Pb2+(aq) +2e- &ruldhar; Pb(s) -0.13
Sn2+(aq) +2e- &ruldhar; Sn(s) -0.14
Ni2+(aq) +2e- &ruldhar; Ni(s) -0.26
Co2+(aq) +2e- &ruldhar; Co(s) -0.28
Cr3+(aq) + e- &ruldhar; Cr2+(aq) -0.41
Fe2+(aq) +2e- &ruldhar; Fe(s) -0.44
Cr3+(aq) +3e- &ruldhar; Cr(s) -0.74
Zn2+(aq) +2e- &ruldhar; Zn(s) -0.76
Cr2+(aq) +2e- &ruldhar; Cr(s) -0.90
Al3+(aq) 3e- &ruldhar; Al(s) -1.67
Mg2+(aq) +2e- &ruldhar; Mg(s) -2.37
Na+(aq) + e- &ruldhar; Na(s) -2.71
Ca2+(aq) +2e- &ruldhar; Ca(s) -2.84
K+(aq) + e- &ruldhar; K(s) -2.93
Li+(aq) + e- &ruldhar; Li(s) -3.04

Note: These reactions are in equilibrium, they can go either way.

To calculate the standard cell potential, from the standard electrode potentials we do the following:

Eθ cell = Eθ right hand half-cell - Eθ left-hand half cell.

The left-hand half cell is conventionally the oxidising half-cell

For example:

Eθ cell = Eθ Cu2+/Cu - Eθ Zn2+/Zn

= +0.34 - (-0.76) V

= +1.1V

This is the shorthand way of representing an electrochemical cell.

For example:

Daniell cell:

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The continuous vertical line represents the phase boundary. The broken vertical line represents the salt bridge.

We can predict the likelihood of a reaction if two systems in the electrochemical series are linked by cells. Remember, the system which is lower in the series will lose electrons, and the one higher in the series will gain electrons.

For example:

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Cell reaction:

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