SATHEE: Electrochemistry - Result Question 33

Electrochemistry - Result Question 33

####35. Consider the half-cell reduction reaction :

$M n^{2 +} + 2 e^{-} \to M n, E^{\circ} = - 1.18 V$

$M n^{2 +} \to M n^{3 +} + e^{-}, E^{\circ} = - 1.51 V$

The $E^{\circ}$ for the reaction $3 M n^{2 +} \to M n^{0} + 2 M n^{3 +}$ , and possibility of the forward reaction are, respectively

[NEET Kar. 2013]

(a) $- 2.69 V$ and no

(b) $- 4.18 V$ and yes

(d) $+ 2.69 V$ and no

Show Answer

Solution:

(a) $Δ E^{\circ} = E_{red}^{\circ} + E_{oxd}^{\circ} = - 1.81 - 1.51 = - 2.69$ Since $Δ E^{\circ}$ is negative

$∴ Δ G = - n F E^{\circ}, Δ G$ will have positive value so, forward reaction is not possible.

The value of electrode potential of a half-cell reaction does not depend on stoichiometric coefficient. e.g.

$\begin{aligned} M n^{2 +} ⟶ M n^{3 +} + e^{-}; E^{\circ} = - 1.51 V \\ 2 M n^{2 +} ⟶ 2 M n^{3 +} + 2 e^{-}; E^{\circ} = - 1.51 V \end{aligned}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Electrochemistry - Result Question 33

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu