θ = 25°C [Cu2+] = [CuO22–] = 1·10-4 mol/ℓ (7): 2CuO22-(aq) + 6H+(aq) + 2e– → Cu2O(s) + 3H2O(ℓ) The electrochemical reduc

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answerhappygod
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θ = 25°C [Cu2+] = [CuO22–] = 1·10-4 mol/ℓ (7): 2CuO22-(aq) + 6H+(aq) + 2e– → Cu2O(s) + 3H2O(ℓ) The electrochemical reduc

Post by answerhappygod »

θ = 25°C [Cu2+] = [CuO22–] = 1·10-4 mol/ℓ
(7): 2CuO22-(aq) + 6H+(aq) + 2e– → Cu2O(s) + 3H2O(ℓ)
The electrochemical reduction potential for reaction (7)
are linear functions in pH and can therefore be written down on the
form
(8) V = a · pH + b
Calculate the constant a of the reaction equation (7)
Needs to be in V
This is the answer i just need an explanation on how to solve
it
 1
1 (6.06 KiB) Viewed 14 times
A) -0.33 V XB) -0.18 V C) -0.28 V D) -0.56 V
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