SATHEE: Electrochemistry - Result Question 47

Electrochemistry - Result Question 47

####49. On the basis of the information available from the reaction

$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 k J m o l^{- 1}$ of $O_{2}$ . The minimum e.m.f required to carry out an electrolysis of $A l_{2} O_{3}$ is $(F = 96500 C m o l^{- 1})$

(a) $8.56 V$

(b) $2.14 V$

(d) $6.42 V$

[2003]

Show Answer

Solution:

(b) $Δ G = - n E F, A l \to A l^{3 +} + 3 e^{-}$

For $1 m o l$ of $A l, n = 3$

$∴$ for $\frac{4}{3} m o l$ of $A l, n = 3 \times \frac{4}{3} = 4$

According to question,

$827 \times 1000 = 4 \times E \times 96500$

$E = \frac{827 \times 1000}{4 \times 96500} = 2.14 V$

Electrochemistry - Result Question 47

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Electrochemistry - Result Question 47

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu