19918
What is the amount of chlorine evolved when \(2 \) amperes of current is passed for \(30 \) minutes in an aqueous solution of \(NaCl\) ............. \(\mathrm{g}\)
19919
On passing a current through \(KCl\) solution, \(19.5\,g\) of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ \(\mathrm{g}\)
1 \(4.5\)
2 \(9\)
3 \(13.5\)
4 \(27\)
Explanation:
\(K^++ e^-= K\) \(Al^{3+} +3 e^- = Al\) Aluminium requires 3 times the amount of electricity to deposit one mole Moles of \(K\) deposited \(=19.5 / 39=0.5\) Moles of \(Al\) deposited \(=0.5 / 3=1 / 6\) Mass of \(Al\) deposited \(=27 \times 1 / 6\) \(=4.5 g\)
ELECTROCHEMISTRY
19920
Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to
1 \(m\,\, \propto {I^2}\)
2 \(m\,\, \propto Q\)
3 \(m\,\, \propto \,\,{Q^2}\)
4 None of these
Explanation:
(b)It is Faraday’s law.
ELECTROCHEMISTRY
19958
In order to separate oxygen from one mole of \({H_2}O\) the required quantity of coulomb would be
1 \(1.93 \times {10^5}\)
2 \(9.6 \times {10^4}\)
3 \(1.8\)
4 \(3.2\)
Explanation:
we have \(1\, mole\) of \(H _2 O\) \(1\, mole\) of \(H _2 O\) given one atom \(O\) \(H _2 O \rightarrow H _2+1 / 2 O _2\) charge on \(O =-2\) electricity required on oxidation of mole of \(H _2 O\) and \(O _2= nf\) where \(n =2\) \(=2 \times 96487\, coulomb\) \(=192974\, coulomb\) \(=1.93 \times 10^5\)
19918
What is the amount of chlorine evolved when \(2 \) amperes of current is passed for \(30 \) minutes in an aqueous solution of \(NaCl\) ............. \(\mathrm{g}\)
19919
On passing a current through \(KCl\) solution, \(19.5\,g\) of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ \(\mathrm{g}\)
1 \(4.5\)
2 \(9\)
3 \(13.5\)
4 \(27\)
Explanation:
\(K^++ e^-= K\) \(Al^{3+} +3 e^- = Al\) Aluminium requires 3 times the amount of electricity to deposit one mole Moles of \(K\) deposited \(=19.5 / 39=0.5\) Moles of \(Al\) deposited \(=0.5 / 3=1 / 6\) Mass of \(Al\) deposited \(=27 \times 1 / 6\) \(=4.5 g\)
ELECTROCHEMISTRY
19920
Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to
1 \(m\,\, \propto {I^2}\)
2 \(m\,\, \propto Q\)
3 \(m\,\, \propto \,\,{Q^2}\)
4 None of these
Explanation:
(b)It is Faraday’s law.
ELECTROCHEMISTRY
19958
In order to separate oxygen from one mole of \({H_2}O\) the required quantity of coulomb would be
1 \(1.93 \times {10^5}\)
2 \(9.6 \times {10^4}\)
3 \(1.8\)
4 \(3.2\)
Explanation:
we have \(1\, mole\) of \(H _2 O\) \(1\, mole\) of \(H _2 O\) given one atom \(O\) \(H _2 O \rightarrow H _2+1 / 2 O _2\) charge on \(O =-2\) electricity required on oxidation of mole of \(H _2 O\) and \(O _2= nf\) where \(n =2\) \(=2 \times 96487\, coulomb\) \(=192974\, coulomb\) \(=1.93 \times 10^5\)
19918
What is the amount of chlorine evolved when \(2 \) amperes of current is passed for \(30 \) minutes in an aqueous solution of \(NaCl\) ............. \(\mathrm{g}\)
19919
On passing a current through \(KCl\) solution, \(19.5\,g\) of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ \(\mathrm{g}\)
1 \(4.5\)
2 \(9\)
3 \(13.5\)
4 \(27\)
Explanation:
\(K^++ e^-= K\) \(Al^{3+} +3 e^- = Al\) Aluminium requires 3 times the amount of electricity to deposit one mole Moles of \(K\) deposited \(=19.5 / 39=0.5\) Moles of \(Al\) deposited \(=0.5 / 3=1 / 6\) Mass of \(Al\) deposited \(=27 \times 1 / 6\) \(=4.5 g\)
ELECTROCHEMISTRY
19920
Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to
1 \(m\,\, \propto {I^2}\)
2 \(m\,\, \propto Q\)
3 \(m\,\, \propto \,\,{Q^2}\)
4 None of these
Explanation:
(b)It is Faraday’s law.
ELECTROCHEMISTRY
19958
In order to separate oxygen from one mole of \({H_2}O\) the required quantity of coulomb would be
1 \(1.93 \times {10^5}\)
2 \(9.6 \times {10^4}\)
3 \(1.8\)
4 \(3.2\)
Explanation:
we have \(1\, mole\) of \(H _2 O\) \(1\, mole\) of \(H _2 O\) given one atom \(O\) \(H _2 O \rightarrow H _2+1 / 2 O _2\) charge on \(O =-2\) electricity required on oxidation of mole of \(H _2 O\) and \(O _2= nf\) where \(n =2\) \(=2 \times 96487\, coulomb\) \(=192974\, coulomb\) \(=1.93 \times 10^5\)
19918
What is the amount of chlorine evolved when \(2 \) amperes of current is passed for \(30 \) minutes in an aqueous solution of \(NaCl\) ............. \(\mathrm{g}\)
19919
On passing a current through \(KCl\) solution, \(19.5\,g\) of potassium is deposited. If the same quantity of electricity is passed through a solution of aluminium chloride, the amount of aluminium deposited is ............ \(\mathrm{g}\)
1 \(4.5\)
2 \(9\)
3 \(13.5\)
4 \(27\)
Explanation:
\(K^++ e^-= K\) \(Al^{3+} +3 e^- = Al\) Aluminium requires 3 times the amount of electricity to deposit one mole Moles of \(K\) deposited \(=19.5 / 39=0.5\) Moles of \(Al\) deposited \(=0.5 / 3=1 / 6\) Mass of \(Al\) deposited \(=27 \times 1 / 6\) \(=4.5 g\)
ELECTROCHEMISTRY
19920
Electrolysis rules of Faraday’s states that mass deposited on electrode is proportional to
1 \(m\,\, \propto {I^2}\)
2 \(m\,\, \propto Q\)
3 \(m\,\, \propto \,\,{Q^2}\)
4 None of these
Explanation:
(b)It is Faraday’s law.
ELECTROCHEMISTRY
19958
In order to separate oxygen from one mole of \({H_2}O\) the required quantity of coulomb would be
1 \(1.93 \times {10^5}\)
2 \(9.6 \times {10^4}\)
3 \(1.8\)
4 \(3.2\)
Explanation:
we have \(1\, mole\) of \(H _2 O\) \(1\, mole\) of \(H _2 O\) given one atom \(O\) \(H _2 O \rightarrow H _2+1 / 2 O _2\) charge on \(O =-2\) electricity required on oxidation of mole of \(H _2 O\) and \(O _2= nf\) where \(n =2\) \(=2 \times 96487\, coulomb\) \(=192974\, coulomb\) \(=1.93 \times 10^5\)
