The number of coulombs required to deposit $5.4 g$ of Aluminium when the given electrode reaction is represented as $A l^{3 +} + 3 e^{-} \to A l$ .

1.83 \times 10^{5} C

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57900 C

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5.86 \times 10^{5} C

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3 F

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Solution

The correct option is D $57900 C$
According to Faraday's law of electrolysis

$W = \frac{Z \times Q}{F}$

$W =$ mass deposited $= 5.4 g$

$Z =$ Equivalent mass $= \frac{M o l a r m a s s}{e^{-} a c c e p t e d}$

$F =$ Faraday constant $= 96500 C$

$Q =$ Charge

$5.4 g = (\frac{27 g}{3}) \times \frac{Q}{96500 C}$

$Q = 57900 C$

Thus, $57900 C$ is required to deposit $5.4 g$ of $A l$

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