The amount of electricity required to deposit 0-9gof aluminum the electrode reaction is Al-3- - 3e- - Al -atomic mass of Al - 27-9-65 - 10-3C4-34 - 10-5C1-93 - 10-4C9-65 - 10-4C

The correct option is A 9.65 × 10^3CAl^3+ + 3e^ → Al 27g of Al mequire = 3 × 96500C ∴ 0.9g of Al requires =3 × 96500/27× 0.9 = 9650C. ...

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Faraday's First Law

The amount of...

Question

The amount of electricity required to deposit 0.9gof aluminum the electrode reaction is
$A l^{3 +} + 3 e^{-} \to A l$
(atomic mass of Al = 27)

9.65 \times 10^{3} C

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1.93 \times 10^{4} C

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9.65 \times 10^{4} C

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4.34 \times 10^{5} C

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