Question

A galvanic cell is based on the following half-reactions at 281 K:

Cu⁺ + e^- → Cu     E^o = 0.521 V
H₂O₂ (aq) + 2 H⁺ + 2 e^- → 2 H₂O     E^o = 1.78 V

What will the potential of this cell be when [Cu⁺] = 0.567 M, [H⁺] = 0.00333 M, and [H₂O₂] = 0.861 M?

E_non

Answer 1

The equilibrium with the more positive E value will move to right .

so reaction at anode(oxidation reaction)

2Cu2Cu⁺ +2e^- E_ox=-0.521V

so reaction at cathode(reduction reaction)

H₂O₂ +2H⁺ +2e^-2H₂O Ered=1.78V

so net reaction:2Cu+H₂O₂+2H⁺2H₂O+2Cu⁺

now according to nernst equation

E=E-(0.0592/2)log[Cu⁺]²/[H⁺]²[H₂O₂] [n=number of electrons exchanged=2]

putting the concentration of Cu⁺,H⁺,H₂O₂

=1.259-1.98408log(.321489)/(1.10889*10^-5)(.861)

=1.124992675V

[E=E_red+E_ox=1.78V-.521V=1.259V]