Standard electrode potential (data page)

The data below tabulates <a href="/facts/Standard_electrode_potential/YQUhbI6K">standard electrode potentials</a> (E°), in <a href="/facts/Volt/KF18MXP2">volts</a> relative to the <a href="/facts/Standard_hydrogen_electrode/c6mksrr1">standard hydrogen electrode</a> (SHE), at:

<ul><li><a href="/facts/Temperature/QyGe4gmj">Temperature</a> 298.15 K (25.00 °C; 77.00 °F);</li>
<li><a href="/facts/Activity_(chemistry)/kGUf85XI">Effective concentration</a> (activity) 1 mol/L for each aqueous or <a href="/facts/Amalgam_(chemistry)/5dRXJyt3">amalgamated</a> (mercury-alloyed) species;</li>
<li>Unit <a href="/facts/Activity_(chemistry)/kGUf85XI">activity</a> for each solvent and pure solid or liquid species; and</li>
<li><a href="/facts/Fugacity/C2gdkThT">Absolute partial pressure</a> 101.325 kPa (1.00000 atm; 1.01325 bar) for each gaseous reagent — the convention in most literature data but not the current <a href="/facts/Standard_state/adrN3C0a">standard state</a> (100 kPa).</li></ul>
Variations from these ideal conditions affect measured voltage via the <a href="/facts/Nernst_equation/bCMCjfqx">Nernst equation</a>.
Electrode potentials of successive elementary half-reactions cannot be directly added. However, the corresponding <a href="/facts/Gibbs_free_energy/bvenDuzh">Gibbs free energy</a> changes (∆G°) must satisfy

∆G° = –zFE°,
where z electrons are transferred, and the <a href="/facts/Faraday_constant/8AXVov0J">Faraday constant</a> F is the <a href="/facts/Conversion_factor/BY9Az4t6">conversion factor</a> describing <a href="/facts/Coulomb/WNV9ueGr">Coulombs</a> transferred per mole electrons. Those Gibbs free energy changes can be added.
For example, from Fe2+ + 2 e− ⇌ Fe(s) (–0.44 V), the energy to form one neutral atom of Fe(s) from one Fe2+ ion and two electrons is 2 × 0.44 eV = 0.88 eV, or 84 907 J/(mol  e−). That value is also the <a href="/facts/Standard_Gibbs_free_energy_of_formation/2LE9C8fl">standard formation energy</a> (∆Gf°) for an Fe2+ ion, since  e− and Fe(s) both have zero formation energy.
Data from different sources may cause table inconsistencies. For example: 
 
 
 
 
 
 
 
 
 
 
 Cu
 
 +
 
 
 +
 
 e
 
 −

⇌

Cu
                  
                    (
                    s
                    )
                  
                
              
              
                
                
                
                  E
                  
                    1
                  
                
                =
                +
                0.520
                
                   V
                
              
            
            
              
              
                
                  
                    Cu
                    
                      2
                      +
                    
                  
                  +
                  2
                  
                  
                    e
                    
                      −

⇌

Cu
                  
                    (
                    s
                    )
                  
                
              
              
                
                
                
                  E
                  
                    2
                  
                
                =
                +
                0.337
                
                   V
                
              
            
            
              
              
                
                  
                    Cu
                    
                      2
                      +
                    
                  
                  +
                  
                    e
                    
                      −

⇌

Cu
 
 +
 
 
 
 
 
 
 
 
 E
 
 3
 
 
 =
 +
 0.159
 
  V
 
 
 
 
 
 
 
 {\displaystyle {\begin{alignedat}{4}&{\ce {Cu+ + e-}}&{}\rightleftharpoons {}&{\ce {Cu(s)}}&\quad \quad E_{1}=+0.520{\text{ V}}\\&{\ce {Cu^2+ + 2e-}}&{}\rightleftharpoons {}&{\ce {Cu(s)}}&\quad \quad E_{2}=+0.337{\text{ V}}\\&{\ce {Cu^2+ + e-}}&{}\rightleftharpoons {}&{\ce {Cu+}}&\quad \quad E_{3}=+0.159{\text{ V}}\end{alignedat}}}
 
 From additivity of Gibbs energies, one must have 
 
 
 
 2
 ⋅
 
 E
 
 2
 
 
 =
 1
 ⋅
 
 E
 
 1
 
 
 +
 1
 ⋅
 
 E
 
 3
 
 
 
 
 {\displaystyle 2\cdot E_{2}=1\cdot E_{1}+1\cdot E_{3}}
 
 But that equation does not hold exactly with the cited values.

Standard electrode potential (data page) open-in-new

Standard electrode potential (data page)