J.R. S. answered 11/26/19
Ph.D. University Professor with 10+ years Tutoring Experience
pH = 2.87
[H3O+] = 1x10-2.87 = 1.35x10-3 M
K = 0.5/(1.35x10-3)2
K = 2.74x105
E = Eº - 0.059/2 log K = 0 - 0.0295 log 2.74x105
E = -0.0295 x 5.44
E = -0.16V
Zoey M.
asked 11/26/19Compute the half-cell potential for reduction of aqueous hydronium ions to hydrogen gas at 0.5 bar and 25 C in buffer solution at pH = 2.87.
J.R. S. answered 11/26/19
Ph.D. University Professor with 10+ years Tutoring Experience
pH = 2.87
[H3O+] = 1x10-2.87 = 1.35x10-3 M
K = 0.5/(1.35x10-3)2
K = 2.74x105
E = Eº - 0.059/2 log K = 0 - 0.0295 log 2.74x105
E = -0.0295 x 5.44
E = -0.16V
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