Tag: applications of electrolysis

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Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $\left[ Ni{ \left( { NO } _{ 3 } \right)  } _{ 2 } \right]$ land chromium nitrate $\left[Cr{ \left( { NO } _{ 3 } \right)  } _{ 3 } \right]$ respectively. If $0.3g$ of nickel was deposited in the first cell, the common of chromium deposited is :
$(at. Wt. Of Ni=59, at. Wt. Of Cr=52)$

  1. $0.1g$
  2. $0.17g$
  3. $0.3g$
  4. $0.6g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Faraday's laws of electrolysis, the mass deposited is proportional to the equivalent weight. Equivalent weight of Ni is 59/2 = 29.5, and for Cr is 52/3 = 17.33. Mass of Cr = (Mass of Ni * Eq. Wt. Cr) / Eq. Wt. Ni = (0.3 * 17.33) / 29.5 = 0.176g.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

A certain quantity of electricity when passed through solution of ${ AgNO } _{ 3 }$, ${ ZnSO } _{ 4 }$, ${ CrI } _{ 3 }$. If X moles of Cr are deposited at its cathode, how many moles of Ag and Zn are deposited at their respective cathodes.

  1. X, X

  2. 3X, 2X

  3. 3X, 1.5X

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Ag^++e^- \longrightarrow Ag$

$Zn^{2+}+2e^- \longrightarrow Zn$
$Cr^{3+}+3e^-\longrightarrow Cr$
Given that $x$ moles of $Cr$ is deposited at it's cathode means in case of $Ag$ it will be $3X$  and for $Zn$ it will be $\cfrac {3X}{2}$ moles deposits at cathode.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

How long (approximate) should water be electrolysed by passing through $100$ amperes current so that the oxygen realised can completely burn $27.66\ g$ of diborane?


(Atomic weight of $B=10.8\ u$ )

  1. $0.8$ hours
  2. $3.2$ hours
  3. $1.6$ hours
  4. $6.4$ hours
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The combustion of diborane (B2H6 + 3O2 -> B2O3 + 3H2O) requires 3 moles of O2 per mole of B2H6. 27.66g of B2H6 is 1 mole (MW=27.66). Thus, 3 moles of O2 are needed. Electrolysis of water (2H2O -> 2H2 + O2) requires 4 moles of electrons per mole of O2. Total charge Q = 3 * 4 * 96500 C. Time = Q / I = 1158000 / 100 = 11580 seconds, which is approximately 3.2 hours.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of Aluminium metal from bauxite by the Hall process is :

  1. 270 kg

  2. 540 kg

  3. 90 kg

  4. 180 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the Hall-Heroult process, aluminum is produced along with the consumption of carbon anodes which react with oxygen to form carbon dioxide according to the reaction 2Al2O3 + 3C -> 4Al + 3CO2. Stoichiometric calculations for 270 kg of Al give 90 kg of carbon consumed.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $[Ni(NO _{3}) _{2}$] and chromium nitrate $[Cr(NO _{3}) _{3}$] respectively.If 0.3 g of nickel was deposited in the first cell, the amount of chromium deposited is:
(at.wt of Ni=59, at. wt. of Cr=52)

  1. 0.1 g

  2. 0.17 g

  3. 0.3 g

  4. 0.6 g
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

On electrolysing a solution of dilute ${ H } _{ 2 }{ SO } _{ 4 }$ between platinum electrodes, the gas evolved at the anode is 

  1. ${ SO } _{ 2 }$
  2. ${ SO } _{ 3 }$
  3. ${ O } _{ 2 }$
  4. ${ H } _{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

During the electrolysis of dilute sulfuric acid using platinum electrodes, water is oxidized at the anode rather than sulfate ions, releasing oxygen gas (O2).

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

In the manufacture of $NaOH$ by the electrolysis of $NaCl$ solution, the cathode and anode are separated using a diaphragm because :

  1. it prevents the reaction betweeen ${H _2}$ and $C{l _2}$ formed
  2. it prevents the mixing of $NaOH$ and $NaC{l _2}$
  3. it prevents the reaction betweeen $Na$ and $C{l _2}$ formed
  4. it increases the yield of $NaOH$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Charge required for liberating $710 g$ of $Cl _{2}(g)$ by electrolyzing a concentrated solution of $NaCl$ will be:

  1. $1.93$ x $10^{5}$ $C$
  2. $1.93$ x $10^{6}$ $C$
  3. $9.65$ x $10^{6}$ $C$
  4. $9.65$ x $10^{5}$ $C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reaction taking place at anode is given by:

$2Cl^- \rightarrow Cl _2+2e^-$
Thus, $2$ moles of $e^-$ are required to liberate $1$ mole of $Cl _2$
Moles of $Cl _2=\dfrac{710}{71}=10$
Hence moles of $e^-$ required $=2\times10=20$
Hence $Q=20F$
$\Rightarrow Q=20\times 96500$
$\Rightarrow Q=1.93\times10^6 C$

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Which process occurs in the electrolysis of an aqueous solution of nickel chloride at nickel anode?

  1. $Ni\rightarrow Ni^{2+}+2e^{-}$
  2. $Ni^{2+}+2e^{-}\rightarrow Ni$
  3. $2CI^{-}\rightarrow 2CI _{2}+2e^{-}$
  4. $2H^{+}+2e^{-}\rightarrow H _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At a nickel anode in an electrolytic cell, the metal itself undergoes oxidation (Ni -> Ni2+ + 2e-) because nickel is an active electrode.