Eo = Eo
R – 0 = Eo
R
The measured emf of the cell:
Pt(s) ç H2(g, 1 bar) ç H
+ (aq, 1 M) çç Cu
2+ (aq, 1 M) ú Cu
is 0 34 V and it is also the value for the standard electrode potential
of the half-cell corresponding to the reaction:
Cu
2+ (aq, 1M) + 2 e
–  ®  Cu(s)
Similarly, the measured emf of the cell:
Pt(s) ç H2(g, 1 bar) ç H
+ (aq, 1 M) çç Zn
2+ (aq, 1M) ç Zn
is -0 76 V corresponding to the standard electrode potential of the
half-cell reaction:
Zn
2+ (aq, 1 M) + 2e
– ®  Zn(s)
The positive  value of the standard electrode potential in the first
case indicates that Cu
2+ ions get reduced more easily than H
+ ions The
reverse process cannot occur, that is, hydrogen ions cannot oxidise Cu
(or alternatively we can say that hydrogen gas can reduce copper ion)
under the standard conditions described above