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# Answer to Question #216287 in Inorganic Chemistry for Nick

Question #216287

Given that Eo for the reaction below is +1.51 V,

(a)       calculate the reduction potential, E, in a solution of pH 2.5 and in which the ratio [Mn2+]:[MnO4-] = 1: 100.

(b)       Show that E = +1:25 V when pH ¼ 3.0 and the ratio [Mn2+]:[MnO4-] =1:100.

(c)       For a ratio [Mn2+]:[MnO4-]=1000 : 1, what must the pH of the solution be to give a value of E = +1.45 V?

(d)       For a ratio [[Mn2+]: [MnO4-]=1 : 100, determine E in a solution of pH 1.8.

1
2021-07-13T02:42:56-0400

we have,

MnO4​ + 8H+ + 5e→ Mn2+ + 4H2​O

we know that

putting Eo=+1.51

.............................1

a)

pH= 2.5 taking antilog we get

[H+]=3.16 x 10-3

[Mn2+]:[MnO4-] = 1: 100.

putting in !1 we get,

b) To show E=1.25 V

pH= 3 taking antilog we get

[H+]=

[Mn2+]:[MnO4-] = 1: 100.

putting in !1 we get,

APPROX

c)

pH= ?

[H+]=?

E = +1.45 V

[Mn2+]:[MnO4-] = 1000 : 1

putting in !1 we get,

d)

pH= 1.8 taking antilog we get

[H+]=1.5 X 10-2

[Mn2+]:[MnO4-] = 1: 100.

putting in !1 we get,

V

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Assignment Expert
14.07.21, 12:33

Dear Nick ,

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Nick
13.07.21, 16:45

Thank you so much for the feedback it definitely helped me a lot. I appreciate your response to the question.