276065
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
1 $1.93 \mathrm{~A}$
2 $9.65 \mathrm{~A}$
3 $19.3 \mathrm{~A}$
4 $0.965 \mathrm{~A}$
Explanation:
No. of moles of $\mathrm{H}_{2}=\frac{1.12}{22400}$ No. of equivalence of hydrogen $=\frac{1.12 \times 2}{22400}=10^{-4}$ No. of Faradays required $=10^{-4}$ $\therefore$ Current to be passed in one second $\begin{aligned} & =96500 \times 10^{-4} \\ & =9.65 \mathrm{~A} \end{aligned}$
Karnataka-CET
ELECTROCHEMISTRY
276067
1 C electricity deposits :
1 half of electrochemical equivalent of $\mathrm{Ag}$
276068
The number of moles of electrons passed when current of $2 \mathrm{~A}$ is passed through a solution of electrolyte for $\mathbf{2 0}$ minutes is
$\mathrm{Q}=\mathrm{I} \times \mathrm{t}=2 \mathrm{~A} \times 20 \times 60 \mathrm{Sec}=2400 \mathrm{C}$ 1 mole of electrons $=96500 \mathrm{C}$ Numbers of mole of electron $=\frac{Q}{96500}$ $\begin{aligned} & =\frac{2400}{96500}=0.02487 \\ & =2.487 \times 10^{-2} \mathrm{~mol} \mathrm{e}^{-} \end{aligned}$
MHT CET-2018
ELECTROCHEMISTRY
276069
How many Faradays of electricity are required to deposit $10 \mathrm{~g}$ of calcium from molten calcium chloride using inert electrodes? $\text { (Molar mass of calcium }=40 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }$
1 $0.5 \mathrm{~F}$
2 $1 \mathrm{~F}$
3 $0.25 \mathrm{~F}$
4 $2 \mathrm{~F}$
Explanation:
$\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}$ Number of moles of $\mathrm{Ca}=\frac{10 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$ 1 mol Ca requires 2 moles of electrons $=2$ faradays of electricity. 0.25 mole $\mathrm{Ca}=2 \times 0.25=0.50$ moles of electrons $=0.5$ faraday's of electricity
MHT CET-2016
ELECTROCHEMISTRY
276071
The amount of silver deposited on passing $2 \mathrm{~F}$ of electricity through aqueous solution of $\operatorname{AgNO}_{3}$
1 $54 \mathrm{~g}$
2 $108 \mathrm{~g}$
3 $216 \mathrm{~g}$
4 $324 \mathrm{~g}$
Explanation:
Number of electron transferred $=1$ Electricity passed $=2 \mathrm{~F}$ Number of moles $=2$ moles Amount of Ag deposited $=$ moles $\times$ molecular weight $\text { of Ag } =2 \times 108=216 \mathrm{~g}$
276065
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
1 $1.93 \mathrm{~A}$
2 $9.65 \mathrm{~A}$
3 $19.3 \mathrm{~A}$
4 $0.965 \mathrm{~A}$
Explanation:
No. of moles of $\mathrm{H}_{2}=\frac{1.12}{22400}$ No. of equivalence of hydrogen $=\frac{1.12 \times 2}{22400}=10^{-4}$ No. of Faradays required $=10^{-4}$ $\therefore$ Current to be passed in one second $\begin{aligned} & =96500 \times 10^{-4} \\ & =9.65 \mathrm{~A} \end{aligned}$
Karnataka-CET
ELECTROCHEMISTRY
276067
1 C electricity deposits :
1 half of electrochemical equivalent of $\mathrm{Ag}$
276068
The number of moles of electrons passed when current of $2 \mathrm{~A}$ is passed through a solution of electrolyte for $\mathbf{2 0}$ minutes is
$\mathrm{Q}=\mathrm{I} \times \mathrm{t}=2 \mathrm{~A} \times 20 \times 60 \mathrm{Sec}=2400 \mathrm{C}$ 1 mole of electrons $=96500 \mathrm{C}$ Numbers of mole of electron $=\frac{Q}{96500}$ $\begin{aligned} & =\frac{2400}{96500}=0.02487 \\ & =2.487 \times 10^{-2} \mathrm{~mol} \mathrm{e}^{-} \end{aligned}$
MHT CET-2018
ELECTROCHEMISTRY
276069
How many Faradays of electricity are required to deposit $10 \mathrm{~g}$ of calcium from molten calcium chloride using inert electrodes? $\text { (Molar mass of calcium }=40 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }$
1 $0.5 \mathrm{~F}$
2 $1 \mathrm{~F}$
3 $0.25 \mathrm{~F}$
4 $2 \mathrm{~F}$
Explanation:
$\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}$ Number of moles of $\mathrm{Ca}=\frac{10 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$ 1 mol Ca requires 2 moles of electrons $=2$ faradays of electricity. 0.25 mole $\mathrm{Ca}=2 \times 0.25=0.50$ moles of electrons $=0.5$ faraday's of electricity
MHT CET-2016
ELECTROCHEMISTRY
276071
The amount of silver deposited on passing $2 \mathrm{~F}$ of electricity through aqueous solution of $\operatorname{AgNO}_{3}$
1 $54 \mathrm{~g}$
2 $108 \mathrm{~g}$
3 $216 \mathrm{~g}$
4 $324 \mathrm{~g}$
Explanation:
Number of electron transferred $=1$ Electricity passed $=2 \mathrm{~F}$ Number of moles $=2$ moles Amount of Ag deposited $=$ moles $\times$ molecular weight $\text { of Ag } =2 \times 108=216 \mathrm{~g}$
276065
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
1 $1.93 \mathrm{~A}$
2 $9.65 \mathrm{~A}$
3 $19.3 \mathrm{~A}$
4 $0.965 \mathrm{~A}$
Explanation:
No. of moles of $\mathrm{H}_{2}=\frac{1.12}{22400}$ No. of equivalence of hydrogen $=\frac{1.12 \times 2}{22400}=10^{-4}$ No. of Faradays required $=10^{-4}$ $\therefore$ Current to be passed in one second $\begin{aligned} & =96500 \times 10^{-4} \\ & =9.65 \mathrm{~A} \end{aligned}$
Karnataka-CET
ELECTROCHEMISTRY
276067
1 C electricity deposits :
1 half of electrochemical equivalent of $\mathrm{Ag}$
276068
The number of moles of electrons passed when current of $2 \mathrm{~A}$ is passed through a solution of electrolyte for $\mathbf{2 0}$ minutes is
$\mathrm{Q}=\mathrm{I} \times \mathrm{t}=2 \mathrm{~A} \times 20 \times 60 \mathrm{Sec}=2400 \mathrm{C}$ 1 mole of electrons $=96500 \mathrm{C}$ Numbers of mole of electron $=\frac{Q}{96500}$ $\begin{aligned} & =\frac{2400}{96500}=0.02487 \\ & =2.487 \times 10^{-2} \mathrm{~mol} \mathrm{e}^{-} \end{aligned}$
MHT CET-2018
ELECTROCHEMISTRY
276069
How many Faradays of electricity are required to deposit $10 \mathrm{~g}$ of calcium from molten calcium chloride using inert electrodes? $\text { (Molar mass of calcium }=40 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }$
1 $0.5 \mathrm{~F}$
2 $1 \mathrm{~F}$
3 $0.25 \mathrm{~F}$
4 $2 \mathrm{~F}$
Explanation:
$\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}$ Number of moles of $\mathrm{Ca}=\frac{10 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$ 1 mol Ca requires 2 moles of electrons $=2$ faradays of electricity. 0.25 mole $\mathrm{Ca}=2 \times 0.25=0.50$ moles of electrons $=0.5$ faraday's of electricity
MHT CET-2016
ELECTROCHEMISTRY
276071
The amount of silver deposited on passing $2 \mathrm{~F}$ of electricity through aqueous solution of $\operatorname{AgNO}_{3}$
1 $54 \mathrm{~g}$
2 $108 \mathrm{~g}$
3 $216 \mathrm{~g}$
4 $324 \mathrm{~g}$
Explanation:
Number of electron transferred $=1$ Electricity passed $=2 \mathrm{~F}$ Number of moles $=2$ moles Amount of Ag deposited $=$ moles $\times$ molecular weight $\text { of Ag } =2 \times 108=216 \mathrm{~g}$
276065
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
1 $1.93 \mathrm{~A}$
2 $9.65 \mathrm{~A}$
3 $19.3 \mathrm{~A}$
4 $0.965 \mathrm{~A}$
Explanation:
No. of moles of $\mathrm{H}_{2}=\frac{1.12}{22400}$ No. of equivalence of hydrogen $=\frac{1.12 \times 2}{22400}=10^{-4}$ No. of Faradays required $=10^{-4}$ $\therefore$ Current to be passed in one second $\begin{aligned} & =96500 \times 10^{-4} \\ & =9.65 \mathrm{~A} \end{aligned}$
Karnataka-CET
ELECTROCHEMISTRY
276067
1 C electricity deposits :
1 half of electrochemical equivalent of $\mathrm{Ag}$
276068
The number of moles of electrons passed when current of $2 \mathrm{~A}$ is passed through a solution of electrolyte for $\mathbf{2 0}$ minutes is
$\mathrm{Q}=\mathrm{I} \times \mathrm{t}=2 \mathrm{~A} \times 20 \times 60 \mathrm{Sec}=2400 \mathrm{C}$ 1 mole of electrons $=96500 \mathrm{C}$ Numbers of mole of electron $=\frac{Q}{96500}$ $\begin{aligned} & =\frac{2400}{96500}=0.02487 \\ & =2.487 \times 10^{-2} \mathrm{~mol} \mathrm{e}^{-} \end{aligned}$
MHT CET-2018
ELECTROCHEMISTRY
276069
How many Faradays of electricity are required to deposit $10 \mathrm{~g}$ of calcium from molten calcium chloride using inert electrodes? $\text { (Molar mass of calcium }=40 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }$
1 $0.5 \mathrm{~F}$
2 $1 \mathrm{~F}$
3 $0.25 \mathrm{~F}$
4 $2 \mathrm{~F}$
Explanation:
$\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}$ Number of moles of $\mathrm{Ca}=\frac{10 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$ 1 mol Ca requires 2 moles of electrons $=2$ faradays of electricity. 0.25 mole $\mathrm{Ca}=2 \times 0.25=0.50$ moles of electrons $=0.5$ faraday's of electricity
MHT CET-2016
ELECTROCHEMISTRY
276071
The amount of silver deposited on passing $2 \mathrm{~F}$ of electricity through aqueous solution of $\operatorname{AgNO}_{3}$
1 $54 \mathrm{~g}$
2 $108 \mathrm{~g}$
3 $216 \mathrm{~g}$
4 $324 \mathrm{~g}$
Explanation:
Number of electron transferred $=1$ Electricity passed $=2 \mathrm{~F}$ Number of moles $=2$ moles Amount of Ag deposited $=$ moles $\times$ molecular weight $\text { of Ag } =2 \times 108=216 \mathrm{~g}$
276065
In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under STP condition. The current to be passed is
1 $1.93 \mathrm{~A}$
2 $9.65 \mathrm{~A}$
3 $19.3 \mathrm{~A}$
4 $0.965 \mathrm{~A}$
Explanation:
No. of moles of $\mathrm{H}_{2}=\frac{1.12}{22400}$ No. of equivalence of hydrogen $=\frac{1.12 \times 2}{22400}=10^{-4}$ No. of Faradays required $=10^{-4}$ $\therefore$ Current to be passed in one second $\begin{aligned} & =96500 \times 10^{-4} \\ & =9.65 \mathrm{~A} \end{aligned}$
Karnataka-CET
ELECTROCHEMISTRY
276067
1 C electricity deposits :
1 half of electrochemical equivalent of $\mathrm{Ag}$
276068
The number of moles of electrons passed when current of $2 \mathrm{~A}$ is passed through a solution of electrolyte for $\mathbf{2 0}$ minutes is
$\mathrm{Q}=\mathrm{I} \times \mathrm{t}=2 \mathrm{~A} \times 20 \times 60 \mathrm{Sec}=2400 \mathrm{C}$ 1 mole of electrons $=96500 \mathrm{C}$ Numbers of mole of electron $=\frac{Q}{96500}$ $\begin{aligned} & =\frac{2400}{96500}=0.02487 \\ & =2.487 \times 10^{-2} \mathrm{~mol} \mathrm{e}^{-} \end{aligned}$
MHT CET-2018
ELECTROCHEMISTRY
276069
How many Faradays of electricity are required to deposit $10 \mathrm{~g}$ of calcium from molten calcium chloride using inert electrodes? $\text { (Molar mass of calcium }=40 \mathrm{~g} \mathrm{~mol}^{-1} \text { ) }$
1 $0.5 \mathrm{~F}$
2 $1 \mathrm{~F}$
3 $0.25 \mathrm{~F}$
4 $2 \mathrm{~F}$
Explanation:
$\mathrm{Ca}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Ca}$ Number of moles of $\mathrm{Ca}=\frac{10 \mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}=0.25 \mathrm{~mol}$ 1 mol Ca requires 2 moles of electrons $=2$ faradays of electricity. 0.25 mole $\mathrm{Ca}=2 \times 0.25=0.50$ moles of electrons $=0.5$ faraday's of electricity
MHT CET-2016
ELECTROCHEMISTRY
276071
The amount of silver deposited on passing $2 \mathrm{~F}$ of electricity through aqueous solution of $\operatorname{AgNO}_{3}$
1 $54 \mathrm{~g}$
2 $108 \mathrm{~g}$
3 $216 \mathrm{~g}$
4 $324 \mathrm{~g}$
Explanation:
Number of electron transferred $=1$ Electricity passed $=2 \mathrm{~F}$ Number of moles $=2$ moles Amount of Ag deposited $=$ moles $\times$ molecular weight $\text { of Ag } =2 \times 108=216 \mathrm{~g}$
