
Match list I with list II
List I List II A) Electrolysis of $Na_2SO_3$ using $Pt$ electrodes 1) 1 Faraday B) The charge carried by $6.023 \times 10^23$ electrons is 2) 3 Faradays C) The amount of electricity required to deposit 27 gms of $Al$ at cathode from molten $Al_2O_3$ is 3) ${H}_{2}(g)/Pt $ D) A gas in contact with an electrode 4) $O_2$ at anode$H_2$ at cathode
A. A-2, B-3, C-4, D-1
B. A-4, B-1, C-2, D-3
C. A-3, B-2, C-4, D-1
D. A-4, B-3, C-2, D-1
| List I | List II |
| A) Electrolysis of $Na_2SO_3$ using $Pt$ electrodes | 1) 1 Faraday |
| B) The charge carried by $6.023 \times 10^23$ electrons is | 2) 3 Faradays |
| C) The amount of electricity required to deposit 27 gms of $Al$ at cathode from molten $Al_2O_3$ is | 3) ${H}_{2}(g)/Pt $ |
| D) A gas in contact with an electrode | 4) $O_2$ at anode$H_2$ at cathode |
Answer
533.1k+ views
Hint: Faraday’s law states that mass deposited by electrolysis, is directly proportional to the current passed through a time of T. 1 mole of any substance have $6.023\times {{10}^{23}}$ atoms. Platinum electrode is called an inert electrode.
Complete answer:
The List I is matched with List II as:
A) Electrolysis of aq.$N{{a}_{2}}S{{O}_{4}}$using Pt electrodes, this will yield ${{O}_{2}}$at the anode and ${{H}_{2}}$at the cathode. This is due to the fact that hydrogen${{H}^{+}}$ion has greater electrode potential than the sodium ion$N{{a}^{+}}$, so Hydrogen ions will move towards the cathode, being positive ions (cations), so${{H}_{2}}$ at the cathode. While at anode ${{O}_{2}}$gas is evolved, as the sulphate ion $S{{O}_{4}}^{2-}$ has less mobility, as sulphur is already in its maximum oxidation state, so no further oxidation, hence ${{O}_{2}}$at the anode.
B) The charge carried by $6.023\times {{10}^{23}}$electrons, is the charge of 1 mole of electrons which is 1 Faraday.
C) The amount of electricity required to deposit 27 gram of aluminum at cathode from molten$A{{l}_{2}}{{O}_{3}}$will be calculated by first law of electrolysis that has mass directly proportional to current and time, as W = $z\times I\times T$ where z is the equivalent weight which is atomic mass upon charge. So, $W=\dfrac{at.wt.}{n\times F}\times q$ where current into time is charge, q. so,
$27=\dfrac{27}{3F}\times q$, So, q = 3F
Hence, 3 F is the correct match.
D. A gas in contact with an inert electrode is the hydrogen gas. An inert electrode is the standard hydrogen electrode SHE, which uses platinum electrodes, and has hydrogen gas ${{H}_{2}}$. Hence ${{H}_{2}}$(g)/ Pt electrode is right.
Thus, the matching order is A-4, B-1, C-2, D-3, so the correct option is B.
Note:
Equivalent weight of any element is the atomic mass upon charge. Here aluminum in option C has equivalent weight calculated as, $\dfrac{23}{3}$ it has a charge of 3 due to 3 electrons in the valence shell. Electrode potential is the ${{E}^{0}}$ values that decide the oxidizing and reducing potential of ions.
Complete answer:
The List I is matched with List II as:
A) Electrolysis of aq.$N{{a}_{2}}S{{O}_{4}}$using Pt electrodes, this will yield ${{O}_{2}}$at the anode and ${{H}_{2}}$at the cathode. This is due to the fact that hydrogen${{H}^{+}}$ion has greater electrode potential than the sodium ion$N{{a}^{+}}$, so Hydrogen ions will move towards the cathode, being positive ions (cations), so${{H}_{2}}$ at the cathode. While at anode ${{O}_{2}}$gas is evolved, as the sulphate ion $S{{O}_{4}}^{2-}$ has less mobility, as sulphur is already in its maximum oxidation state, so no further oxidation, hence ${{O}_{2}}$at the anode.
B) The charge carried by $6.023\times {{10}^{23}}$electrons, is the charge of 1 mole of electrons which is 1 Faraday.
C) The amount of electricity required to deposit 27 gram of aluminum at cathode from molten$A{{l}_{2}}{{O}_{3}}$will be calculated by first law of electrolysis that has mass directly proportional to current and time, as W = $z\times I\times T$ where z is the equivalent weight which is atomic mass upon charge. So, $W=\dfrac{at.wt.}{n\times F}\times q$ where current into time is charge, q. so,
$27=\dfrac{27}{3F}\times q$, So, q = 3F
Hence, 3 F is the correct match.
D. A gas in contact with an inert electrode is the hydrogen gas. An inert electrode is the standard hydrogen electrode SHE, which uses platinum electrodes, and has hydrogen gas ${{H}_{2}}$. Hence ${{H}_{2}}$(g)/ Pt electrode is right.
Thus, the matching order is A-4, B-1, C-2, D-3, so the correct option is B.
Note:
Equivalent weight of any element is the atomic mass upon charge. Here aluminum in option C has equivalent weight calculated as, $\dfrac{23}{3}$ it has a charge of 3 due to 3 electrons in the valence shell. Electrode potential is the ${{E}^{0}}$ values that decide the oxidizing and reducing potential of ions.
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