SATHEE: Electrochemistry - Result Question 22

Electrochemistry - Result Question 22

####23. $A l_{2} O_{3}$ is reduced by electrolysis at low potentials and high currents. If $4.0 \times 10^{4}$ amperes of current is passed through molten $A l_{2} O_{3}$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency and At. mass of $A l = 27 g m o l^{- 1}$ )

[2009]

(a) $8.1 \times 10^{4} g$

(b) $2.4 \times 10^{5} g$

(d) $9.0 \times 10^{3} g$

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Solution:

(a) As $Q = i \times t$

$\begin{aligned} ∴ Q & = 4.0 \times 10^{4} \times 6 \times 60 \times 60 C \\ = 8.64 \times 10^{8} C \end{aligned}$

Now since $96500 C$ liberates $9 g$ of Al

$\begin{aligned} 8.64 \times 10^{8} C liberates \frac{9}{96500} \times 8.64 \times 10^{8} g A l \\ = 8.1 \times 10^{4} g of Al \end{aligned}$

24 .

(d) No. of gram equivalent of $H^{+} =$ No. of eq. of $A l^{3 +} = \frac{4.5}{9} = 0.5$

Eq. wt of $A l^{3 +} = \frac{27}{3} = 9$

No. of gm eq. of $H^{+} =$ no. of mole of $H^{+}$

Hence, Mass of $H^{+} = 0.5 \times 1 g = 0.5 g$

We know that, $2 g H_{2}$ at $S T P = 22.4 L$

$∴ 0.5 g H_{2} at S T P = \frac{22.4}{2}, 0.5 = 5.6 L$

Electrochemistry - Result Question 22

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Electrochemistry - Result Question 22

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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