SATHEE: Redox Reactions And Electrochemistry Ques 727

Redox Reactions And Electrochemistry Ques 727

Question: Given

(i) $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \overset{}{\to}$ $M n^{2 +} + 4 H_{2}, O$ (ii) $M n O_{2} + 4 H^{+} + 2 e^{-} \overset{}{\to}$ $M n^{2 +} + 2 H_{2}, O E^{0} = x_{2} V$ Find $E^{0}$ for the following reaction $M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \overset{}{\to} M n O_{2} + 2 H_{2}, O$

Options:

A) $x_{2} - x_{1}$

B) $x_{1} - x_{2}$

C) $\frac{5 x_{1} - 2 x_{2}}{3}$

D) $\frac{2 x_{1} - 5 x_{2}}{3}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $Δ G_{1}^{0} = - 5 x_{1} F$ ; $Δ G_{2}^{0} = - 2 x_{2} F$ $Δ G_{3}^{0} = Δ G_{1}^{0} - Δ G_{2}^{0}$ ; $Δ G_{3}^{0} = 2 x_{2} F - 5 x_{1} F$

$∴$ $x = \frac{5 x_{1} - 2 x_{2}}{3}$

Redox Reactions And Electrochemistry Ques 727

Question: Given

Options:

Answer:

Solution:

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Redox Reactions And Electrochemistry Ques 727

Question: Given

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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