Given below are the half - cell reactions :

$\begin{matrix} {Mn}^{2 +} + 2 e^{-} \to Mn; & E^{\circ} = - 1.18 V \\ 2 ({Mn}^{3 +} + e^{-} \to {Mn}^{2 +}); & E^{\circ} = + 1.51 V \end{matrix}$

The $E^{\circ}$ for $3 {Mn}^{2 +} \to Mn + 2 {Mn}^{3 +}$ will be :

Question

Given below are the half - cell reactions : Mn 2 + + 2 e - → Mn ; E ∘ = - 1.18 V 2 Mn 3 + + e - → Mn 2 + ; E ∘ = + 1.51 V The E ∘ for 3 Mn 2 + → Mn + 2 Mn 3 + will be :

Aicharya · Accepted Answer

&#916; G &#8728; = - nF E &#8728; Mn + 2 + 2 e - &#8594; Mn ........... (1) &#916; G = - 2 &#215; F &#215; - 1.18 = + 2.36 F 2 Mn + 3 + 2 e - &#8594; 2 Mn + 2 ....... (2) &#916; G = - 2 &#215; F &#215; +1.51 = - 3.02 F (1) - (2) 3 Mn + 2 &#8594; Mn + 2 Mn + 3 &#916; G = + 2.36F + &#160; 3.02F = 5.38 F &#916; G = - 2 &#215; F &#215; E E = 5.38F - 2 &#215; F = - 2.69 V We know that when E cell < 0, cell is non-spontaneous.

Given below are the half - cell reactions : Mn 2 + + 2 e - → Mn ; E ∘ = - 1.18 V 2 Mn 3 + + e - → Mn 2 + ; E…

Solution