19833
The electrochemical equivalent of a metal is \(3.3 \times {10^{ - 7}}\,kg/coulomb\). The mass of the metal liberated at the cathode when a \(3\, A\) current is passed for \(2\) seconds will be
19834
Faraday's 2nd law states that mass deposited on the electrode is directly proportional to
1 Atomic mass
2 Atomic mass $×$ Velocity
3 Atomic mass/Valency
4 Valency
Explanation:
(c) \(m = zq\), \(z\) = atomic mass / valence
ELECTROCHEMISTRY
19835
The relation between Faraday constant \((F)\), chemical equivalent \((E)\) and electrochemical equivalent \((Z)\) is
1 \(F = EZ\)
2 \(F = \frac{Z}{E}\)
3 \(F = \frac{E}{Z}\)
4 \(F = \frac{E}{Z^2}\)
Explanation:
The relation between Faraday constant chemical equivalent and electrochemical equivalent is \(F =\frac{ E }{ Z }\) where \(E\) is chemical equivalent, \(Z\) is electrochemical equivalent and \(F\) is faraday constant.
ELECTROCHEMISTRY
19836
On passing \(96500\) coulomb of charge through a solution \(CuS{O_4}\) the amount of copper liberated is
1 \(64\, gm\)
2 \(32\, gm\)
3 \(32\, kg\)
4 \(64\, kg\)
Explanation:
(b) \(1\) faraday (\(96500\,C\)) is the electricity which liberated that amount of substance which is equal to equivalent wt. So liberated amount of \(Cu\) is \(\frac{{63.5}}{2}\) \( = 31.25\,gm \approx 32\,gm\)
19833
The electrochemical equivalent of a metal is \(3.3 \times {10^{ - 7}}\,kg/coulomb\). The mass of the metal liberated at the cathode when a \(3\, A\) current is passed for \(2\) seconds will be
19834
Faraday's 2nd law states that mass deposited on the electrode is directly proportional to
1 Atomic mass
2 Atomic mass $×$ Velocity
3 Atomic mass/Valency
4 Valency
Explanation:
(c) \(m = zq\), \(z\) = atomic mass / valence
ELECTROCHEMISTRY
19835
The relation between Faraday constant \((F)\), chemical equivalent \((E)\) and electrochemical equivalent \((Z)\) is
1 \(F = EZ\)
2 \(F = \frac{Z}{E}\)
3 \(F = \frac{E}{Z}\)
4 \(F = \frac{E}{Z^2}\)
Explanation:
The relation between Faraday constant chemical equivalent and electrochemical equivalent is \(F =\frac{ E }{ Z }\) where \(E\) is chemical equivalent, \(Z\) is electrochemical equivalent and \(F\) is faraday constant.
ELECTROCHEMISTRY
19836
On passing \(96500\) coulomb of charge through a solution \(CuS{O_4}\) the amount of copper liberated is
1 \(64\, gm\)
2 \(32\, gm\)
3 \(32\, kg\)
4 \(64\, kg\)
Explanation:
(b) \(1\) faraday (\(96500\,C\)) is the electricity which liberated that amount of substance which is equal to equivalent wt. So liberated amount of \(Cu\) is \(\frac{{63.5}}{2}\) \( = 31.25\,gm \approx 32\,gm\)
19833
The electrochemical equivalent of a metal is \(3.3 \times {10^{ - 7}}\,kg/coulomb\). The mass of the metal liberated at the cathode when a \(3\, A\) current is passed for \(2\) seconds will be
19834
Faraday's 2nd law states that mass deposited on the electrode is directly proportional to
1 Atomic mass
2 Atomic mass $×$ Velocity
3 Atomic mass/Valency
4 Valency
Explanation:
(c) \(m = zq\), \(z\) = atomic mass / valence
ELECTROCHEMISTRY
19835
The relation between Faraday constant \((F)\), chemical equivalent \((E)\) and electrochemical equivalent \((Z)\) is
1 \(F = EZ\)
2 \(F = \frac{Z}{E}\)
3 \(F = \frac{E}{Z}\)
4 \(F = \frac{E}{Z^2}\)
Explanation:
The relation between Faraday constant chemical equivalent and electrochemical equivalent is \(F =\frac{ E }{ Z }\) where \(E\) is chemical equivalent, \(Z\) is electrochemical equivalent and \(F\) is faraday constant.
ELECTROCHEMISTRY
19836
On passing \(96500\) coulomb of charge through a solution \(CuS{O_4}\) the amount of copper liberated is
1 \(64\, gm\)
2 \(32\, gm\)
3 \(32\, kg\)
4 \(64\, kg\)
Explanation:
(b) \(1\) faraday (\(96500\,C\)) is the electricity which liberated that amount of substance which is equal to equivalent wt. So liberated amount of \(Cu\) is \(\frac{{63.5}}{2}\) \( = 31.25\,gm \approx 32\,gm\)
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ELECTROCHEMISTRY
19833
The electrochemical equivalent of a metal is \(3.3 \times {10^{ - 7}}\,kg/coulomb\). The mass of the metal liberated at the cathode when a \(3\, A\) current is passed for \(2\) seconds will be
19834
Faraday's 2nd law states that mass deposited on the electrode is directly proportional to
1 Atomic mass
2 Atomic mass $×$ Velocity
3 Atomic mass/Valency
4 Valency
Explanation:
(c) \(m = zq\), \(z\) = atomic mass / valence
ELECTROCHEMISTRY
19835
The relation between Faraday constant \((F)\), chemical equivalent \((E)\) and electrochemical equivalent \((Z)\) is
1 \(F = EZ\)
2 \(F = \frac{Z}{E}\)
3 \(F = \frac{E}{Z}\)
4 \(F = \frac{E}{Z^2}\)
Explanation:
The relation between Faraday constant chemical equivalent and electrochemical equivalent is \(F =\frac{ E }{ Z }\) where \(E\) is chemical equivalent, \(Z\) is electrochemical equivalent and \(F\) is faraday constant.
ELECTROCHEMISTRY
19836
On passing \(96500\) coulomb of charge through a solution \(CuS{O_4}\) the amount of copper liberated is
1 \(64\, gm\)
2 \(32\, gm\)
3 \(32\, kg\)
4 \(64\, kg\)
Explanation:
(b) \(1\) faraday (\(96500\,C\)) is the electricity which liberated that amount of substance which is equal to equivalent wt. So liberated amount of \(Cu\) is \(\frac{{63.5}}{2}\) \( = 31.25\,gm \approx 32\,gm\)
