Question #216287

Given that E^{o} 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 [Mn_{2}^{+}]:[MnO_{4}^{-}] = 1: 100.

(b) Show that E = +1:25 V when pH ¼ 3.0 and the ratio [Mn_{2}^{+}]:[MnO_{4}^{-}] =1:100.

(c) For a ratio [Mn_{2}^{+}]:[MnO_{4}^{-}]=1000 : 1, what must the pH of the solution be to give a value of E = +1.45 V?

(d) For a ratio [[Mn_{2}^{+}]: [MnO_{4}^{-}]=1 : 100, determine E in a solution of pH 1.8.

Expert's answer

we have,

** MnO**_{4}^{−}** + 8H**^{+ }**+ 5e**^{−}**→ Mn2**^{+ }**+ 4H**_{2}**O**

we know that

putting E^{o}=+1.51

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

**a)**

pH= 2.5 taking antilog we get

[H+]=3.16 x 10^{-3}

[Mn_{2}^{+}]:[MnO_{4}^{-}] = 1: 100.

putting in !1 we get,

**b) To show E=1.25 V**

pH= 3 taking antilog we get

[H+]=

[Mn_{2}^{+}]:[MnO_{4}^{-}] = 1: 100.

putting in !1 we get,

^{APPROX}

**c)**

pH= ?

[H+]=?

E = +1.45 V

[Mn_{2}^{+}]:[MnO_{4}^{-}] = 1000 : 1

putting in !1 we get,

** **

**d)**

pH= 1.8 taking antilog we get

[H+]=1.5 X 10^{-2}

[Mn_{2}^{+}]:[MnO_{4}^{-}] = 1: 100.

putting in !1 we get,

V

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Assignment Expert14.07.21, 12:33Dear Nick ,

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Nick13.07.21, 16:45Thank you so much for the feedback it definitely helped me a lot. I appreciate your response to the question.

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