Question

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 [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.

Answer 1

we have,

MnO₄⁻​ + 8H⁺ + 5e⁻→ Mn2⁺ + 4H₂​O

we know that

$E=E^o-{0.059 \over 5}log{[Mn^{2+}]\over [MnO_4^{-}][H^{+}]^8}$

putting E^o=+1.51

$E=1.51-{0.059 \over 5}log{[Mn^{2+}]\over [MnO_4^{-}][H^{+}]^8}$ .............................1

a)

pH= 2.5 taking antilog we get

[H+]=3.16 x 10^-3

[Mn₂⁺]:[MnO₄^-] = 1: 100.

putting in !1 we get,

$E=1.51-{0.059 \over 5}log{[1]\over [100][3.16 \times10^{-3}]^8}$

$E=1.51-{0.06\over5} \times18$

$E=1.294$

b) To show E=1.25 V

pH= 3 taking antilog we get

[H+]= $10^{-3}$

[Mn₂⁺]:[MnO₄^-] = 1: 100.

putting in !1 we get,

$E=1.51-{0.059 \over 5}log{[1]\over [100][10^{-3}]^8}$

$E=1.51-{0.059 \over 5}\times22$

$E=1.25 V$ ^APPROX

c)

pH= ?

[H+]=?

E = +1.45 V

[Mn₂⁺]:[MnO₄^-] = 1000 : 1

putting in !1 we get,

$1.45=1.51-{0.059 \over 5}log{[1000]\over [1][H^{+}]^8}$

$0.06={0.06\over5}log{[1000]\over [1][H^{+}]^8}$

$10^5={1\over 1000 H^+}$

$[H^+]=10^{-7}$

d)

pH= 1.8 taking antilog we get

[H+]=1.5 X 10^-2

[Mn₂⁺]:[MnO₄^-] = 1: 100.

putting in !1 we get,

$E=1.51-{0.059 \over 5}log{[1]\over [100][1.5\times 10^{-2}]^8}$

$E=1.51-{0.059\over 5}{(14-1.6)}$

$E=1.361$ V