(B) The reduction reaction is: $Fe^{3+} + 3e^{-} \longrightarrow Fe$From the stoichiometry,$3 \ mol$ of electrons are required to deposit $1 \ mol$ of

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(B) The reduction reaction is: $Fe^{3+} + 3e^{-} \longrightarrow Fe$From the stoichiometry,$3 \ mol$ of electrons are required to deposit $1 \ mol$ of $Fe$ $(56 \ g)$.Total charge required for $0.3482 \ g$ of $Fe$ is:$Q = \frac{3 \times 96500 \times 0.3482}{56} \approx 1800 \ C$Using the formula $Q = I \times t$ (where $t$ is in seconds):$1800 = 1.5 \times t$$t = \frac{1800}{1.5} = 1200 \ s$Converting time to minutes:$x = \frac{1200}{60} = 20 \ min$

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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$A$ solution of $Fe_{2}(SO_{4})_{3}$ is electrolyzed for '$x$' min with a current of $1.5 \ A$ to deposit $0.3482 \ g$ of $Fe$. The value of $x$ is $.......$. [nearest integer]
Given : $1 \ F = 96500 \ C \ mol^{-1}$
Atomic mass of $Fe = 56 \ g \ mol^{-1}$

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Faraday’s law of electrolysis

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Calculate the quantity of electricity required to liberate $0.1 \ mole$ of chlorine gas during electrolysis of molten sodium chloride. (in $C$)

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The number of moles of electrons required to deposit $36 \ g$ of $Al$ from an aqueous solution of $Al(NO_3)_3$ is (At. wt. of $Al = 27$)

DifficultAP EAMCET 2012

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Aluminium oxide may be electrolysed at $1000 \, ^\circ C$ to furnish aluminium metal (At. Mass $= 27 \, amu$; $1 \, F = 96,500 \, C$). The cathode reaction is $Al^{3+} + 3e^- \to Al^0$. To prepare $5.12 \, kg$ of aluminium metal by this method would require:

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$A$ solution of $Fe_{2}(SO_{4})_{3}$ is electrolyzed for '$x$' min with a current of $1.5 \ A$ to deposit $0.3482 \ g$ of $Fe$. The value of $x$ is $.......$. [nearest integer]Given : $1 \ F = 96500 \ C \ mol^{-1}$Atomic mass of $Fe = 56 \ g \ mol^{-1}$

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$A$ current of $3 \ A$ is passed through a molten calcium salt for $1 \ hr \ 47 \ min \ 13 \ s$. The mass of calcium deposited is $(Molar \ mass \ of \ Ca = 40 \ g \ mol^{-1})$ (in $g$)

Calculate the quantity of electricity required to liberate $0.1 \ mole$ of chlorine gas during electrolysis of molten sodium chloride. (in $C$)

The unit of Faraday is

The number of moles of electrons required to deposit $36 \ g$ of $Al$ from an aqueous solution of $Al(NO_3)_3$ is (At. wt. of $Al = 27$)

Aluminium oxide may be electrolysed at $1000 \, ^\circ C$ to furnish aluminium metal (At. Mass $= 27 \, amu$; $1 \, F = 96,500 \, C$). The cathode reaction is $Al^{3+} + 3e^- \to Al^0$. To prepare $5.12 \, kg$ of aluminium metal by this method would require:

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$A$ solution of $Fe_{2}(SO_{4})_{3}$ is electrolyzed for '$x$' min with a current of $1.5 \ A$ to deposit $0.3482 \ g$ of $Fe$. The value of $x$ is $.......$. [nearest integer]
Given : $1 \ F = 96500 \ C \ mol^{-1}$
Atomic mass of $Fe = 56 \ g \ mol^{-1}$