Question

Find Oxidation Potential of Cr(s)/Cr3+(0.1M) at 25 degree Celsius. If E not of Cr3+/Cr= 0.75v?​

Answer 1

Answer:

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Answer 2

Answer:

-0.73v

Explanation:

E=E* - 0.059 log10[p] / n x [R]

E=0.75 - 0.059 log10[1] / 3 x [0.1]

E=0.75 - 0.06 log10^10/3

E=0.75 - 0.06 x 1 /3

E=0.75 - 0.02

E(reduction potential) =0.73v

E(oxidation potential) = -0.73v

(0.059 is taken as 0.06 as approximate value)