Electrochemistry

InText Solution

Question 1: How would you determine the standard electrode potential of the system Mg²⁺|Mg?

Answer: We know that it is not possible to measure the electrode potential of a half cell. We can only measure the difference between electrode potentials of two half cells, which gives the cell potential of the cell. For measuring the standard electrode potential of the given system, we need to use Mg as one electrode and a standard hydrogen electrode as the second electrode. Since, electrode potential of standard hydrogen electrode is zero, value of cell potential will give the value of standard electrode potential of the system Mg²⁺|Mg.

Question 2: Can you store copper sulphate solutions in a zinc pot?

Answer: We know that zinc is more reactive metal than copper. Hence, zinc will displace copper from copper sulphate solution. So, keeping copper sulphate solution in a zinc pot will result in copper sulphate turning to zinc sulphate and the zinc pot getting corroded.

Question 3: Consult the table of standard electrode potentials and suggest three substances that can oxidize ferrous ions under suitable conditions.

Answer: Ferrous ion oxidizes to give ferric ions as follows:

`Fe^(2+)→Fe^(3+)+e^(-)`

Here, `E^Θ=-0.77V`

It is clear that a substance with reduction potential greater than 0.77 V can oxidize ferrous ion into ferric ion. Such elements are Br₂, Cl₂ and F₂.

Question 4: Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

Answer: For hydrogen electrode `H^(+)+e^(-)→12H_2`

Using Nernst Equation

`E_(H|1/2H_2)``=E_(H|1/2H_2)°-(0.0591)/n\lo\g\1/([H^+])`

`=0-(0.0591)/1lo\g\1/(10^(-10))`

`=-0.0591xx10=-0.591 V`

Question 5: Calculate the emf of the cell in which the following reaction takes place:

`Ni(s)+2Ag^(+)(0.002M)→Ni^(2+)(0.160M)+2Ag(s)`

Answer: Using Nernst equation

`E_(ce\ll)=E_(ce\ll)°-(0.0591)/n\l\og ([Ni^(2+)])/([Ag^(+)]^2)`

`=1.05V-(0.0591)/2l\og (0.160)/((0.002)^2)`

`=1.05-(0.0591)/2lo\g(4xx10^4)`

`=1.05-(0.0591)/2(4.6021)`

`=1.05-0.14 V=0.91 V`

Question 6: The cell in which the following reaction occurs:

`2Fe^(3+)(aq)+2I^(-)(aq)→2Fe^(2+)(aq)+I_2(s)` has `E_(ce\ll)^Θ=0.236V` at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer: For this cell, `n=2`

`Δ_TG°=-nF\E_(ce\ll)°`

`=-2xx96500xx0.236=-45.55` kJ mol^-1

Moreover, `Δ_TG°=-2.303RT lo\g K_C`

Or, `lo\gK_C=(-ΔG°)/(2.303RT)`

`=(-45.55)/(2.303xx8.314xx10^(-3)xx298)=7.983`

Or, `K_C=an\ti\log(7.983)=9.616xx10^7`

Question 7: Why does the conductivity of a solution decrease with dilution?

Answer: We know that conductivity of a solution is the conductance of ions per unit volume of the solution. Number of ions per unit volume of solution decreases on dilution. Due to this, conductivity of a solution decreases with dilution.

Question 8: Suggest a way to determine the Λ^o_m value of water.

Answer: The value of Λ°_m can determined by using Kohlrausch’s law as folllows:

`Λ_m°= Λ_m°(HC\l)+ Λ_m°(Na\OH)- Λ_m°(Na\Cl)`

Now, `Λ_m°` values of HCl, NaOH and NaCl are substituted in above equation to find the value of `Λ_m°` of water.

Question 9: The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate the degree of dissociation and dissociation constant. Given &lambda^o(H⁺) = 349.6 S cm² mol^-1 and λ°(HCCO^-) = 54.6 S cm² mol^-1.

Answer: `Λ_m°(HCOOH)=λ°(H^(+))+λ°(HCOO^(-))`

`=349.6+54.6=404.2` S cm² mol^-1

Given, `Λ_m^C=46.1` S cm² mol^-1

So, `α=(Λ_m&C)/(Λ_m°)`

`=(46.1)/(404.2)=0.114`

`HC\OO\H⇄HC\OO^(-)+H^(+)`

Initial concentration: c ⇄ 0 + 0

Concentration at equilibrium: c(1 - α) ⇄ cα + cα

So, `K_α=(cαcα)/(c(1-α)=(cα^2)/(1-α)`

`=(0.025xx(0.114)^2)/(1-0.114)=3.67xx10^(-4)`

Question 10: If a current of 0.5 ampere flows through a metallic wire for 2 hours then how many electrons would flow through the wire?

Answer: Given current = 0.6 A and time = 2 hours

We know charge can be calculated as follows:

`Q=It`

`=0.5xx(2xx60xx60)=3600 C`

We know `1F ≈96500C` per mol

Moreover, 1 mol of electrons = 6.02 `xx` 10²³ electrons

Hence, number of electrons

`=(6.23xx10^(23))/(96500)xx3600``=2.246xx10^(22)` electrons

Question 11: Suggest a list of metals that are extracted electrolytically.

Answer: Sodium, calcium, magnesium, aluminium

Question 12: Consider the reaction: `Cr_2O_7^(2-)+14H^(+)+6e^(-)``→2Cr^(3+)+7H_2O`

What is the quantity of electricity in coulombs needed to reduce 1 mol of `Cr_2O_7^(2-)`?

Answer: From the reaction, it is clear that 1 mole of Cr₂O₇^2- needs 6F of electrical charge

`6F=6xx96500=579000 C`

Question 13: Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.

Answer: Following reactions take place while recharging the lead storage battery

At anode: `Pb\SO_4(s)+H^(+)(aq)+2e^(-)``→Pb(s)+HS\O_4^(-)(aq)`

At cathode: `Pb\SO_4(s)+2H_2O(l)``→Pb\O_2(s)+ HS\O_4^(-)(aq)+3H^(+)(aq)+2e^(-)`

Overall Reaction:

`2Pb\SO_4(s)+2H_2O(l)``→Pb(s)+Pb\O_2(s)+2H^(+)(aq)+2HS\O_4^(-)(aq)`

Question 14: Suggest two materials other than hydrogen that can be used as fuels in fuel cells.

Answer: Methane and Methanol

Question 15: Explain how rusting of iron is envisaged as setting up of an electrochemical cell.

Answer: The chemistry of corrosion is quite complex but it can be considered as an electrochemical phenomenon. Oxidation takes place at a particular spot of iron object, and that spot behaves as anode. The reaction at anode can be written as follows:

Anode: `2Fe(s)→2Fe^(2+)+4e^(-)`

`E_((Fe^(2+))/(Fe))^Θ=-0.44V`

Electrons released at the anodic spot move through the metal and go to another spot on the metal. These electrons reduce oxygen in the presence of H⁺ . This spot behaves as cathode where following reaction takes place.

Cathode: `O_2(g)+4H^(+)(aq)+4e^(-)→2H_2O(l)`

`E^Θ=1.23V`

Overall Reaction:

`2Fe(s)+O_2(g)+4H^(+)(aq)→2Fe^(2+)(aq)+2H_2O(l)`

`E_((cell))^Θ=1.67 V`

Ferrous ions are further oxidized by atmospheric oxygen to ferric ions and come out as rust in the form of hydrated ferric oxide (Fe₂O₃.`x`H₂O)