Electrolysis and its applications - class-XII

Adding water increases the volume, which shifts the equilibrium. According to Le Chatelier's principle, the system will try to increase the number of ions. However, the dilution effect (concentration = moles/volume) dominates, leading to a decrease in concentration.

KNO3 is a strong electrolyte where both K+ and NO3- ions have high reduction/oxidation potentials compared to water. During electrolysis, water is preferentially electrolyzed, leaving the concentration of K+ and NO3- ions effectively unchanged.

Calculate theoretical yield using Faraday's laws: Q = I * t = 19.3 * 7200 = 138960 C. Moles of electrons = Q / F = 138960 / 96500 = 1.44 mol. Reaction: Mn2+ + 2H2O -> MnO2 + 4H+ + 2e-. Moles of MnO2 = 1.44 / 2 = 0.72 mol. Mass = 0.72 * 87 = 62.64 g. Efficiency = (52.2 / 62.64) * 100 = 83.33%.

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.

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$.

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.
$4OH^- \rightarrow 2H _2O + O _2 \uparrow + 4e^-$

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.

At the anode, oxidation occurs. Nickel metal (the anode) loses electrons to form Ni2+ ions, which dissolve into the solution.

at cathode:
$2H^+ +2e^- \rightarrow H _2$
at anode:
$2 H _2O \rightarrow O _2 + 4H^+ + 4 e^-$

None of the above ions discharge at anode ; instead Ni atoms of the anode lose electrons to form $Ni^{2+}$ ions.

It is HCl having unique property which does not conduct electricity when it is in gas phase since the gas phase of HCl is in the form of molecules and not in ionic form, the anhydrous HCl gas does not conduct electricity. But HCl solution or Aq HCl conducts electricity since HCl completely dissociates in solution to form Hydronium $H _3O^+$ and $Cl^-$ ions which are getting transported into opposite direction when potential electric junction is created by applying differential voltage, thus migration of these ions in opposite directions to each other allows electricity and thus becomes conductor of electricity.

At cathode:
$Cu^{+2} +2e^- \rightarrow Cu$
At anode:
$2 H _2O \rightarrow O _2 + 4H^+ + 4 e^-$

During the electrolysis of aqueous zinc sulphate, Zn2+ ions are reduced at the cathode to form zinc metal (Zn).

$NH _3$ doesnt conducts electricity in gaseous form but in solution form due to formation of $NH _4^+,  OH^-$ ions, it conducts electricity.

Correction : in question sodium cannot be obtained. In aqueous solution , it is $H _2O$ which is reduced.
$2H _2O + 2e^{-} \rightarrow 2H _2 + 2\overset {\ominus}{O} H$

$CaCl _2$ is added to lower the melting point from 1080 K (melting point of NaCl) to 850 K. This is due to following reasons.
a. At such high temperature, it is very difficult to maintain Na in the molten state during electrolysis.
b. Moreover, Na is voltile at this temperature and a part of it would vapourize.
c. Molten Na, gets dispersed in molten NaCl to form metallic fog at high temperature.
d. Both Na and $Cl _2$ produced in the electrolysis have a corrosive action on the vessel used.
The over all reactions nare as follows :
At cathode : $Na^{\oplus} + e^{-} \rightarrow Na$
At anode : $Cl^{\ominus} \rightarrow \displaystyle \frac{1}{2} Cl _2 + e^{-}$

Molten $Al _2O _3 + Na _3AiF _6 (Cryolite) + CaF _2 (Flurospar) + AlF _3$ is used.
The electrolysis of this mixture is carried out using carbon (graphite) electrodes. The $O _2$ liberated at the anode reacts with the graphite anode needs to be replaced periodically.
Thus, graphite anodes do not have long life.
The overall reactions are as follows :
Cathode : $Al^{+3}(melt) + 3e^{-} \rightarrow Al (l)$.
Anode : $C(s) + O^{2-} (melt) \rightarrow CO(g) + 2e^{-}$
$C(s) + 2O^{2-} (melt) \rightarrow CO _2 (g) + 4e^{-}$

Moissan in 1896 prepared $F _2$ as :
$KHF _2 \rightleftharpoons KF + HF$
$KF \rightleftharpoons  K^{\oplus} + F^{\ominus}$
At anode : $2F^{\ominus} \rightarrow F _2 + 2e^{-}$
At cathode : $K^{\oplus} + e^{-} \rightarrow K$
$K + 2HF \rightarrow 2KF + F _2$

In molten sodium hydride (NaH), electrolysis involves hydride ions (H-) migrating to the anode where they are oxidized to release hydrogen gas. The reason states that hydrogen ions have a lower discharge potential, which is incorrect contextually since NaH contains H- ions acting as the source of hydrogen at the anode, not H+ ions at the cathode in this specific molten setup. Thus, the assertion is true and the reason is false.

Aqueous $CuBr _{2}$ :
At cathode:

$CaH _2, \xrightarrow {Electrolysis}, Ca^{2+}, +, 2H^{-}$

Here,
Electrolysis of $H _2O$ takes place:

Electrolysis of concentrated $H _{2}SO _{4}$ at $0 - 5^{\circ}C$ using Pt electrode produces $H _{2}S _{2}O _{8}$.

During electrolysis of aqueous $CaCl _2$

$E^{\ominus} _{ox}$ of $Cu$ =$ - 0.34 $
Using copper electrodes, copper will oxidize : 
$Cu, \rightarrow, Cu^{2+}, +, 2e^{-}$
So oxygen is not evolved.

Ion with higher potential will go to corresponding electrode and discharge.
So,
$NaF$:
Cathode : $2H^{\oplus}, +, 2e^{-}, \rightarrow, H _{2}$
Anode : $4\overset{\ominus}{O}H, \rightarrow, O _{2}, +, 2H _{2}O, +, 4e^{-}$

Since the reduction potential of $H _2O$ is greater than reduction potential of $Na^{\oplus}\, so\, H _2O$ undergoes reduction to give
$H _2(g)$ at cathode.

$\displaystyle H _2O\, +\, e^-\, \rightarrow\, \overset {\ominus}O H +\, \frac{1}{2} H _2(g)$

Similarly, the oxidation potential of $H _2O$ is greater than the oxidation potential of $SO _4^{2-}$ ion. So, $H _2O$ undergoes oxidation to give $O _2(g)$ at anode.

Sodium acetate undergoes electrolysis to form ethane gas, carbondioxide and hydrogen gas.

In the production of Al (in Hall's process), reactions are:
Cathode:
$Al^{3+}, +, 3e^-, \rightarrow, Al$
Anode:
$2O^{-2}, \rightarrow, O _2, +, 4e^-$
$C, +, O _2, \rightarrow, CO _2$

At cathode:
$\displaystyle H _2O, +, e^-, \rightarrow, \overset{\ominus}O H, +, \frac{1}{2} H _2, (one)$
At anode:
$2HCOO^{\ominus}, \rightarrow, 2CO _2, +, H _2, (two)$
$H _2, and, CO _2$ at anode and $H _2$ at cathode

Upon electrolysis,
At anode:
$2OH^-$ $\rightarrow H _2O + \frac{1}{2}O _2 + 2e^- $
Hence option A is the right answer.

Haber process is used for producing ammonia from nitrogen and hydrogen.
Down's Process. Molten NaCl upon electrolysis will give Na metal and $Cl _2$ gas. Aqueous NaCl will form NaOH,$H _2 , Cl _2$.

If aq. $NaCl$ is electrolyzed using graphite as anode and $Hg$ as cathode then products are $Cl _2$ gas at anode and $Na$ at the cathode.

$Overall\quad reaction :\ { 2H } _{ 2 }O+{ 2Cl }^{ - }+{ 2Na }^{ + }\rightarrow { 2Na }^{ + }+{ 2OH }^{ - }+{ H } _{ 2 }+{ Cl } _{ 2 }$

$NaBr\rightleftharpoons Na^{+} + Br^{-}$
$2H _{2}O + 2e\rightarrow H _{2} + 2OH^{-}$
$Na^{+} + OH^{-} \rightarrow NaOH$ At cathode
$Br^{-}\rightarrow Br + e^{-}$
$Br + Br \rightarrow Br _{2}$ At anode
So the products are $H _{2}$ and $NaOH$ (at cathode) and $Br _{2}$ (at anode).

When water is added, HCl, KOH, NaI and $CaCl _2$ will produce an electrolytic solution. However, $N _2$ will not produce an electrolytic solution on addition of water. Nitrogen is a homo-nuclear diatomic molecule which does not dissociate into aqueous solution.

Electrolysis of aqueous $KCl$ solution will give $KOH$. This is similar to the preparation of $NaOH$ by electrolysis of aqueous $NaCl$ solution. The electrolysis of fused $KOH$ will give potassium metal.

Higher the standard reduction potential, higher will be the tendency to undergo reduction so, more fast will be its deposition on electrode (cathode) .

During the electrolysis of aqueous sodium chloride solution, the products are $NaOH , Cl _2$ and $H _2$.
$NaCl(aq) + H _2O(l) \rightarrow Na^+ (aq) +OH^-(aq) +\frac{1}{2} H _2(g) +\frac{1}{2}Cl _2(g)$
Due to formation of base, it turns res litmus into blue.

$At\ cathode:{ 2H } _{ 2 }O+{ 2e }^{ - }\rightarrow { H } _{ 2 }+{ 2OH }^{ - }\ At\ anode:{ 2Cl }^{ - }\rightarrow { Cl } _{ 2 }+{ 2e }^{ - }$

 The conductivity of NaCl is more in an aqueous solution than in alcohol.
Water is a highly polar solvent. In water, NaCl completely dissociates into sodium ions and chloride ions.
$\displaystyle NaCl \rightarrow Na^+  +  Cl^-$
The electrical conductivity of aqueous NaCl solution is due to a presence of these ions.
Alcohol is less polar than water. Hence, the degree of ionization of NaCl in alcohol is much lower than that in water.   Hence, a lesser number of current carrying ions are present in alcohol. Hence, $NaCl$ in alcohol has lower conductivity than $NaCl$ in an aqueous solution.

As chloride ion have more oxidation potential than hydroxide ion, so $Cl _2$ is liberated at the anode.
And reduction potential of hydrogen ion is more so hydrogen gas produces at cathode so as the current flows, pH of the solution around the cathode increases.

As the reduction potential of $K^+$ and $ Na^+$ are lower than hydrogen ion, hydrogen ion gets preferably reduced and hydrogen gas is produced at the cathode.
Also, as oxidation potential of $OH^-$ is higher than sulphate ion, it will oxidise first and oxygen is evolved at the anode.

Use the Nernst equation for the cell potential: E_cell = E_cell_0 - (0.059/n) * log(Q). The cell reaction involves Tl+ and Pb2+. Use the given potential and concentrations to solve for [Tl+] at equilibrium, then use Ksp = [Tl+][Br-].

a. Reduction potential of $H _{2}O, >$ Reduction potential of $Na^{\oplus}$
Hence, 
Cathode : $2H _{2}O, +, 2e^{-}, \rightarrow, 2 \overset{\ominus}{O}H, +,\begin{matrix} H _{2}\ (solution, is, basic)\end{matrix}$

Anode : $CH _{3}, COO^{\ominus}, \overset{Kolbe's, electrolysis}{\rightarrow}, C _{2}H _{6}, (Ethane), +, 2CO _{2}$

At cathode : Reduction of $Na^{\oplus}$ does not occur but reduction of $H _{2}O$ occurs to give $\overset{\ominus}{O}H$ and $H _{2} (g)$, so pOH decreases and pH increases. Hence, option A is correct

$CuSO _4(aq)\, \xrightarrow{Electrolsis} \, Cu^{2+}(aq)\, +\, SO _4^{2-}(aq)$

At cathode: $Cu^{2+}(aq)\, +\, 2e^-\, \rightarrow\, Cu\, (reduction)$

The blue color of $CuSO _4$ disappears due to the deposition of Cu on Pt electrode.

At anode: $H _2O\, \rightarrow\, 2H^{\oplus}\, +\, 2e^-\, \frac{1}{2} O _2(g)$

Since oxidation potential of $H _2O$ > oxidation potential of $SO _4^{2-}$, so oxidation of $H _2O$ occurs and $O _2(g)$ is evolved at anode.

The colourless solution is due to the formation of $H _2SO _4$ as follows:

Because  ${ SO } _{ 4 }^{ 2- }$ mobility is very less as compared to ${ OH }^{ - }$ . So instead of ${ SO } _{ 4 }^{ 2- },{ OH }^{ - }$ undergo oxidation to form ${ O } _{ 2 }$ at anode. 

Electrolysis of ${ H } _{ 2 }O$ helps in separation of ${ H } _{ 2 }O$ to ${ H } _{ 2 }$ & ${ O } _{ 2 }$