The magnitude of the change in oxidizing power of the $M n O_{4}^{-} / M n^{2 +}$ **couple is x * 10^-4 V, if the H⁺ concentration is decrease from 1M to 10^-4M at 25°C (Assume concentration of $M n O_{4}^{-}$ and Mn²⁺ to be same on change in H⁺ concentration). The value of x is………………… (Rounded off to the nearest integer)

[Given :** $\frac{2.303 R T}{F} = 0.059$ ]

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Vishal Baghel | Contributor-Level 10

5 months ago

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

$E_{1} = E^{0} - \frac{0.059}{5} l o g \frac{[M n^{2 +}]}{[M n O_{4}^{-}]} {[\frac{1}{H^{+}}]}^{8}$

[H⁺] = 1M

$E_{2} = E^{0} - \frac{0.059}{5} l o g \frac{[M n^{2 +}]}{[M n O_{4}^{-}]} + \frac{0.059}{5} l o g 1 0^{- 32}$

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 * 10^-4 V

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The cell potential = _____ ×10⁻² V. (Round off to the Nearest Integer).
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E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
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E_cell = 1.05 - (0.059 / 2) log ( [Ni²? ] / [Ag? ]²)
= 1.05 - (0.059 / 2) log ( [10? ³] / [10? ³]²)
= 1.05 - (0.059 × 3) / 2 = 1.05 - 0.0885 = 0.9615 volt
There is a misprint in the question. The E? _cell is incorrectly given as 10.5 V. This should have been 1.05 volt.