Tag: chemistry

Questions Related to chemistry

A big irregular shaped vessel contained water, conductivity of which was $ 2.56 \times 10^{-3}\, S^{-1} \, m^{-1}.$ 585 g of NaCI was then added to the water and conductivity after the addition of Nacl, was found to be $ 3.06 \times 10^{-3}\,S^{-1}\, m^{-1}. $ The molar conductivity of Nacl at this concentration is $ 1.5 \times 10^{-2}\, S^{-1}\, mol^{-1}.$The capacity of vessel if it is fulfilled with water, is

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. $ 3 \times 10^{4} 1$

  2. $ 30 \,1$

  3. $ 3 \times 10^{8} 1$

  4. $ 3 \times 10^{5} 1 $

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Correct Option: A

The electrolysis of which electrolyte gives the same products in the fused state as well as in the aqueous solution state?

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. NaCl

  2. KCl

  3. AuCl$ _{3}$

  4. BaCl$ _{2}$

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Correct Option: C
Explanation:

All have $Cl^-$ ions, $\therefore  Cl _2$, is released at anode
The element with reduction potential > red. potential of $H^+$ will give same results for aqueous and molter states.
From the given options only $SRP  of  Au^+  >  SRP  of  H^+$
Hence, for other salts in their aqueous solutions $H^+$  gets reduced

Hence, option C is correct.

If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions discharged at cathode are:

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. H$^+$

  2. Na$^+$

  3. OH$^-$

  4. Cl$^-$

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Correct Option: B
Explanation:

In presence of Hg electrode, sodium ions are discharged in the form of a mercury sodium amalgam and chloride ions are converted to chlorine. 
So, $Na^+$ ions discharged at cathode.

Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out:

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. in the presence of NaCl.

  2. in the presence of fluoride.

  3. in the presence of cryolite which forms a melt with lower melting temperature.

  4. in the presence of cryolite which forms a melt with higher melting temperature.

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Correct Option: C
Explanation:

Pure alumina melts at about $ 2000^{0}C$ and is a bad conductor of electricity so fused cryolite and fluorspar is added to lower the melting point and the mixture melts at $ 900^0C$.

State True or False.
Electrolysis of H$ _2$O gives H$ _2$ at the anode.

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. True

  2. False

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Correct Option: B
Explanation:
The given statement is false.
The electrolysis of $H _2O$ is explained below:
Cathode (reduction):
$2 H _2O(l) + 2e^- \rightarrow  H _2(g) + 2 OH^-(aq)$
Anode (oxidation):
$ 4 OH^-(aq) \rightarrow O _2(g) + 2 H _2O(l) + 4 e^-$
Thus the electrolysis of $H _2O$ gives $H _2$ at the cathode.
Electrolysis of CaH$ _2$ gives $H _2$ at anode.
chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. True

  2. False

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Correct Option: A
Explanation:

The above statement is true.
Electrolysis of $CaH _2$,
At anode,
$2H^-\rightarrow H _2+2e$
at cathode
$Ca^{2+} +2e \rightarrow Ca$

State True or False.
Electrolysis of molten $KCl$ gives Cl$ _2$ at cathode.

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. True

  2. False

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Correct Option: A
Explanation:

Electrolysis of molten KCl,
At anode,
$2Cl^-\rightarrow Cl _2+2e$
at cathode
$K^++e\rightarrow K$

Thus the above statement is true.

State True or False.
The ratio of gases liberated at cathode and anode during electrolysis of CH$ _3$COONa(aq) is 1 : 3.

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. True

  2. False

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Correct Option: A
Explanation:

The given statement is True.
 The ratio of gases liberated at cathode and anode during electrolysis of $\displaystyle CH _3COONa$ (aq) is $\displaystyle 1:3$
$\displaystyle 2CH _3COONa \rightarrow CH _3-CH _3 \uparrow + 2CO _2  \uparrow+2NaOH+H _2 \uparrow$
hydrogen gas is liberated at cathode and ethane and carbon dioxide gases are liberated at anode.

Assertion: Dilute H$ _2$SO$ _4$ on electrolysis liberates O$ _2$ at the anode.
Reason: Hydroxide ions have lower discharge potential than sulphate ions.

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. Both Assertion and Reason are true and Reason is the correct explanation of Assertion

  2. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion

  3. Assertion is true but Reason is false

  4. Assertion is false but Reason is true

  5. Both Assertion and Reason are false

Show answer
Correct Option: A
Explanation:

Dilute H$ _2$SO$ _4$ on electrolysis liberates O$ _2$ at the anode. This is due to lower discharge potential of hydroxide ions than sulphate ions.
$4OH^- \rightarrow 2H _2O + O _2 \uparrow + 4e^-$

In the electrolysis of $Na _{2}SO _{4}$ solution using inert electrode

a) the anodic reaction is

$2H _{2}O\rightarrow O _{2}(g)+4e^{-}+4H^{+}$

b)$H _{2}(g)$ and $O _{2}(g)$ is produced in a molar ratio of 2:1

c) 23 grams of sodium is produced at the cathode

d) the cathode reaction is $Na^{+}+e^{-}\rightarrow Na$

chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

  1. a and b are correct

  2. c,d are correct

  3. only c is correct

  4. all are correct

Show answer
Correct Option: A
Explanation:

The discharge potential of $ { H }^{ + } $ ions is lower than $ { Na }^{ + } $ ions. Hence $ { H }^{ + } $ ions will be liberated at cathode in preference to $ { Na }^{ + } $ ions.
The discharge potential of $ { OH }^{ - } $ ions is lower than $ { SO } _{ 4 }^{ 2- } $ ions. Hence $ { OH }^{ - } $ ions will be liberated at anode (in the form of $ { O } _{ 2 } $ molecule) in preference to $ { SO } _{ 4 }^{ 2- } $ ions. This is the electrolysis of water. Hence $ { H } _{ 2 } $ and $ { O } _{ 2 } $ are produced in the molar ratio of 2:1 which is same as that present in water molecule.