(D) Given: $t = 1 \ hr \ 47 \ min \ 13 \ s = (1 \times 3600) + (47 \times 60) + 13 = 3600 + 2820 + 13 = 6433 \ s$ $Molar \ mass \ of \ Ca = 40 \ g \ m

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(D) Given: $t = 1 \ hr \ 47 \ min \ 13 \ s = (1 \times 3600) + (47 \times 60) + 13 = 3600 + 2820 + 13 = 6433 \ s$ $Molar \ mass \ of \ Ca = 40 \ g \ mol^{-1}$ $Current \ (I) = 3 \ A$ $n-factor \ for \ Ca^{2+} + 2e^- \rightarrow Ca \text{ is } 2$. Using Faraday's law of electrolysis: $w = \frac{M \times I \times t}{n \times 96500}$ $w = \frac{40 \times 3 \times 6433}{2 \times 96500}$ $w = \frac{771960}{193000} \approx 4.0 \ g$

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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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 $g$)

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Electrochemistry

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Faraday’s law of electrolysis

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Match List $I$ with List $II$.
List $I$ (Conversion) List $II$ (Number of Faraday required)
$A$. $1 \text{ mole of } H_2O \text{ to } O_2$ $I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$ $II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$ $III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$ $IV$. $5F$

Choose the correct answer from the options given below:

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On passing $C$ ampere of current for time $t$ seconds through $1 \ L$ of $2 \ M$ $CuSO_4$ solution (atomic weight of $Cu = 63.5$), the amount $m$ of $Cu$ (in grams) deposited on the cathode will be:

MediumWBJEE 2012

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The number of $Faradays$ needed to reduce $4 \ g$ equivalents of $Cu^{++}$ to $Cu$ metal will be:

Medium

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The amount of silver deposited by passing $241.25 \ C$ of charge through a silver nitrate solution is

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How many seconds of current at $241.25 \ A$ must be passed to deposit all the silver present in $125 \ mL$ of $1 \ M \ AgNO_3$ solution? $(1 \ F = 96500 \ C)$

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List $I$ (Conversion)	List $II$ (Number of Faraday required)
$A$. $1 \text{ mole of } H_2O \text{ to } O_2$	$I$. $3F$
$B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$	$II$. $2F$
$C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$	$III$. $1F$
$D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$	$IV$. $5F$

$A$ 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 $g$)

Similar Questions

On passing $C$ ampere of current for time $t$ seconds through $1 \ L$ of $2 \ M$ $CuSO_4$ solution (atomic weight of $Cu = 63.5$), the amount $m$ of $Cu$ (in grams) deposited on the cathode will be:

The number of $Faradays$ needed to reduce $4 \ g$ equivalents of $Cu^{++}$ to $Cu$ metal will be:

The amount of silver deposited by passing $241.25 \ C$ of charge through a silver nitrate solution is

How many seconds of current at $241.25 \ A$ must be passed to deposit all the silver present in $125 \ mL$ of $1 \ M \ AgNO_3$ solution? $(1 \ F = 96500 \ C)$

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