Answer Happy

Hey
I have a question where I have to find two constants from
a reaction equation
The equation is at 25 degress and [Cu2+] = [CuO22–] =
1·10-4 mol/ℓ
The electrochemical reduction potential for reactions (4) and
(7) are linear functions in pH and can therefore be written down on
the form
(8) V = a · pH + b. Where a and b have the unit Volt and in the
current situation will be constants
From reaction (4) i have to find conctant a in V
From reaction (7) i have to find conctant b in V
(4) 2Cu²+(aq) + H2O(1) + 2e → Cu2O(s) + 2H+(aq)

) (7) 2Cu022-(aq) + 6H*(aq) + 2e → Cu2O(s) + 3H2O(1) ->