The magnitude of the change in oxidising power of the MnO 4 - / Mn 2 + couple is x × 10 - 4 V if the H + concentration is decreased from 1 M to 10 - 4 M at 25 ° C . (Assume concentration of MnO 4 - an

Question

The magnitude of the change in oxidising power of the ${MnO}_{4}^{-} / {Mn}^{2 +}$ couple is $x \times 10^{- 4} V$ if the $H^{+}$ concentration is decreased from $1 M$ to $10^{- 4} M$ at $25 ° C .$ (Assume concentration of ${MnO}_{4}^{-}$ and ${Mn}^{2 +}$ to be same on change in $H^{+}$ concentration). The value of $x$ is __________.

(Rounded off to the nearest integer)

[Given: $\frac{2.303 RT}{F} = 0.059$ ]

TestHub · Accepted Answer

Eqn is- MnO 4 - + H ⊕ + 5 e - → Mn + 2 + 4 H 2 O Nernst equation: E cell = E Cell 0 - 0 . 059 5 log Mn + 2 MnO 4 - 1 H + 8 (I) Given H ⊕ = 1 M E 1 = E 0 - 0 . 059 5 log Mn + 2 MnO 4 - (II) Now: H ⊕ = 10 - 4 M E 2 = E 0 - 0 . 059 5 log Mn + 2 MnO 4 - × 1 10 - 4 8 = E 0 - 0 . 059 5 log Mn + 2 MnO 4 - + 0 . 059 5 log 10 - 32 therefore: E 1 - E 2 = 0 . 059 5 × 32 = 0 . 3776 V = 3776 × 10 - 4 x = 3776

Chemistry - Electrochemistry Question with Solution | TestHub

Doubts & Discussion