19802
To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is
1 \(1\, ampere\)
2 \(96000\, amperes\)
3 \(96500\, farads\)
4 \(96500\, coulombs\)
Explanation:
(d) \(96500\, coulombs\) of charge is needed to deposit one gram equivalent of an element at an electrode.
ELECTROCHEMISTRY
19803
To deposit one litre of hydrogen at \(22.4\) atmosphere from acidulaled water, the quantity of electricity that must pass through is ............. \(\mathrm{coulomb}\)
1 \(1\)
2 \(22.4\)
3 \(96500\)
4 \(193000\)
Explanation:
(d) \({V_2} = \frac{{22.4 \times 1}}{1} = 22.4\,litre\) at \(NTP\) \(11.2\) litre of \(H_2\) is liberated by \(96,500\,C\) \(22.4\) litre of \(H_2\) is liberated by \(96500 × 2\) = \(1,93,000\,C\)
ELECTROCHEMISTRY
19804
The amount of substance liberated on electrodes during electrolysis when \(1\, coulomb\) of electricity is passed, is
1 Chemical equivalent
2 Electrochemical equivalent
3 Equivalent weight
4 One mol
Explanation:
(b) From \(m = ZQ\); if \(Q = 1\,C\) ==> \(m = Z\)
19802
To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is
1 \(1\, ampere\)
2 \(96000\, amperes\)
3 \(96500\, farads\)
4 \(96500\, coulombs\)
Explanation:
(d) \(96500\, coulombs\) of charge is needed to deposit one gram equivalent of an element at an electrode.
ELECTROCHEMISTRY
19803
To deposit one litre of hydrogen at \(22.4\) atmosphere from acidulaled water, the quantity of electricity that must pass through is ............. \(\mathrm{coulomb}\)
1 \(1\)
2 \(22.4\)
3 \(96500\)
4 \(193000\)
Explanation:
(d) \({V_2} = \frac{{22.4 \times 1}}{1} = 22.4\,litre\) at \(NTP\) \(11.2\) litre of \(H_2\) is liberated by \(96,500\,C\) \(22.4\) litre of \(H_2\) is liberated by \(96500 × 2\) = \(1,93,000\,C\)
ELECTROCHEMISTRY
19804
The amount of substance liberated on electrodes during electrolysis when \(1\, coulomb\) of electricity is passed, is
1 Chemical equivalent
2 Electrochemical equivalent
3 Equivalent weight
4 One mol
Explanation:
(b) From \(m = ZQ\); if \(Q = 1\,C\) ==> \(m = Z\)
19802
To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is
1 \(1\, ampere\)
2 \(96000\, amperes\)
3 \(96500\, farads\)
4 \(96500\, coulombs\)
Explanation:
(d) \(96500\, coulombs\) of charge is needed to deposit one gram equivalent of an element at an electrode.
ELECTROCHEMISTRY
19803
To deposit one litre of hydrogen at \(22.4\) atmosphere from acidulaled water, the quantity of electricity that must pass through is ............. \(\mathrm{coulomb}\)
1 \(1\)
2 \(22.4\)
3 \(96500\)
4 \(193000\)
Explanation:
(d) \({V_2} = \frac{{22.4 \times 1}}{1} = 22.4\,litre\) at \(NTP\) \(11.2\) litre of \(H_2\) is liberated by \(96,500\,C\) \(22.4\) litre of \(H_2\) is liberated by \(96500 × 2\) = \(1,93,000\,C\)
ELECTROCHEMISTRY
19804
The amount of substance liberated on electrodes during electrolysis when \(1\, coulomb\) of electricity is passed, is
1 Chemical equivalent
2 Electrochemical equivalent
3 Equivalent weight
4 One mol
Explanation:
(b) From \(m = ZQ\); if \(Q = 1\,C\) ==> \(m = Z\)
19802
To deposit one gm equivalent of an element at an electrode, the quantity of electricity needed is
1 \(1\, ampere\)
2 \(96000\, amperes\)
3 \(96500\, farads\)
4 \(96500\, coulombs\)
Explanation:
(d) \(96500\, coulombs\) of charge is needed to deposit one gram equivalent of an element at an electrode.
ELECTROCHEMISTRY
19803
To deposit one litre of hydrogen at \(22.4\) atmosphere from acidulaled water, the quantity of electricity that must pass through is ............. \(\mathrm{coulomb}\)
1 \(1\)
2 \(22.4\)
3 \(96500\)
4 \(193000\)
Explanation:
(d) \({V_2} = \frac{{22.4 \times 1}}{1} = 22.4\,litre\) at \(NTP\) \(11.2\) litre of \(H_2\) is liberated by \(96,500\,C\) \(22.4\) litre of \(H_2\) is liberated by \(96500 × 2\) = \(1,93,000\,C\)
ELECTROCHEMISTRY
19804
The amount of substance liberated on electrodes during electrolysis when \(1\, coulomb\) of electricity is passed, is
1 Chemical equivalent
2 Electrochemical equivalent
3 Equivalent weight
4 One mol
Explanation:
(b) From \(m = ZQ\); if \(Q = 1\,C\) ==> \(m = Z\)