CHEMISTRY A LEVEL(FORM SIX) NOTES – ORGANIC CHEMISTRY 1.1 -POLYMERS (2)

I. CONSTRUCTION OF ELECTROCHEMICAL CELLS

Electrochemical cells are devices that use chemical reactions to produce electrical power. These are sometimes known as Galvanic or Voltaic cells.

e.g. dry cells, car batteries.

It contains two half cells connected together by external circuits.

Half cell is an arrangement which consists of an electrode dipped into a solution containing its ions. When the two half cells are connected, the resulting component is called electrochemical power. A good example is Daniel cell. It is constructed from Zinc and copper electrode.

PROCEDURE FOR CONSTRUCTION OF ELECTROCHEMICAL CELL

1. Identify between the electrodes, which electrode is supplying or gaining electrons by studying their electrode potential.
2. An electrode with negative standard electrode potential is more reactive (supplies electrons) than the positive one.
3. If both are negative the one which has more negative electrode potential is more reactive.

E.g. Sn = -0.16

Mg = -0.25 (more reactive)

If both electrodes are positive, the one which has less positive value is more reactive.

E.g. 0.07v (more reactive)

0.2v

The electrode which supplies electrons should be placed on the left and electrode which gains electrons should be on the right hand side.

E.g. construct a Daniel cell and shows the direction of flow of electrons and current given that

The electrode which supplies electrons oxidation takes place and the electrode is oxidized. This electrode is called ANODE. The electrode which receives electrons reduction takes place and the electrode is REDUCED. It is called CATHODE.

Zn – anode

Cu – cathode

Overall reaction is obtained by adding the two half reactions and is called cell reaction.

Anodic reaction:

Cathodic reaction:

Cell reaction:

Electrochemical cell can also be represented in an abbreviation way known as cell notation.

Anode cathode

I.e. for Daniel cell

Questions

1. Given the overall reactions, write their corresponding cell notations
   1. 
   2. 
2. From the cell notation give the cell reactions
   1. 
   2. 

Answers

Cu(aq) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

ELECTROMOTIVE (EMF) FORCE OF A CELL

The difference between electrode potential of the two electrodes constituting an electrochemical cell is known as electromotive (emf) or cell potential.

This acts as a driving force for a cell reaction and it is expressed in volts.

The emf of a cell is calculated by subtracting the standard electrode potential of the left electrode from that of the right electrode.

For Daniel cell

= 0.34 – (-0.76)

= 1.1v

Alternatively

Since standard electrode potentials are in reduced form, for oxidation half reaction the sign of electrode potential should be reversed.

= 0.76 + 0.34

= 1.1v

1. Prediction of the occurrence of chemical reaction.

If the emf of the cell calculated is negative the reaction is non spontaneous i.e. the reaction does not occur in the way it is written unless external forces apply but the reaction is spontaneous in the opposite direction.

If the emf of the cell is positive the reaction is spontaneous.

Example: given the following reaction

Predict the direction of the reaction given that reduction potential of nickel (Ni²⁺/Ni) E^o = -0.25v of mercury Hg⁺/Hg E^o = 0.14v

Solution

Cell notation

The reaction is non spontaneous since emf is negative hence backward reaction is favored.

Example

For each of the following reactions at standard condition, decide if it occurs spontaneously in the direction written.

Given for .

ANSWERS

= -0.14 – (-0.76)

= 0.62V

2. Cell notation:
   

=1.09 – (1.36)

= -0.27V

3. Cell notation:
   

=-0.4 – 0.34

=-0.74V

Therefore reaction (1) is spontaneous while reaction (2) and (3) are non-spontaneous.

Example

Briefly explain what happens when

1. Fe is dipped in CuSO₄ solution
2. Cu is dipped in FeSO₄ solution

Example

Given the following values

a) State which species are the strongest oxidant and which oxidant and weakest reductant.

b) State which species is the strongest oxidant and weakest reductant.

c) The lead rods are placed in a solution of each CuSO₄, FeSO₄, AgNO₃ and ZnSO₄. In which solution do you expect coating of another metal on lead rod? Explain.

ANSWERS

2. (i) Since Fe has a negative reduction potential then it has a positive oxidation potential and hence will displace copper metal from CuSO₄ solution.

(ii) Since Cu has positive reduction potential, then it has a negative oxidation potential and hence it will NOT displace Fe from FeSO₄ solution therefore no reaction will be possible.

3. (a) strongest oxidant

Weakest reductant

Maximum and minimum emf of a cell

It is possible to decide which cell has to be constructed either of great or smallest emf for a given electrodes. A cell with greatest emf is obtained by choosing two electrodes of greatest cell reactivity difference. A cell with minimum emf, choose two electrodes with closest reactivity.

Example

Study the following electrodes.

4. Explain how you can construct a cell that will yield

1. Maximum emf
2. Minimum emf

ANSWER

For maximum emf, we choose two electrodes of greatest cell reactivity difference.

= 2.87 – (-3.04)

= 5.91v

For minimum emf, we choose two electrodes with closest reactivity

= -0.4 – (-0.44)

= 0.04v

EFFECTS OF CONCENTRATION AND TEMPERATURE ON CELL POTENTIALS

We have been considering electrode potential under standard conditions of molar solution, pressure of 1 atm and 298K. When the conditions are altered the values of electrode potential changes thus we have to define the potential of the cell under non-standard conditions.

The Nernst equation shows the relationship between emf of the cell at standard conditions and emf under non-standard conditions.

Where;

– emf of a cell at any conditions

– emf of a cell under standard conditions

R – Universal gas constant (8.314 Jmol^-1K^-1)

n – Number of moles electrons being transferred

F – Faraday’s constant (96500 C)

At standard temperature (298K)

OR

The above equation can be applied to half reactions and overall reactions.

NOTE

Always write a balanced cell reaction to obtain ‘n’ and position of ions, either reactants or products together with their stoichiometry.

Example

Calculate the emf of the given cell at standard temperature.

Given

Alternative Solution 1

= -0.13 – (-2.56)

= 2.43v

Alternative Solution 2

n = 6

Now;

Example 2

Calculate the emf of Daniel cell at using 2M ZnSO₄ solutions and 0.5M CuSO₄ solution

Alternative Solution 1

= 0.34 – (-0.76)

= 1.1v

Alternative Solution 2

Half reactions

(Oxidation)

(Reduction)

Here n = 2

Now using

E_cell = 1.08v

Example 3

Calculate the emf for the following voltaic cells

1. 
2. Zn/Zn²⁺ (10^-3 M)//Ag⁺ (10^-3 M)/Ag
3. 

Given

Alternative Solution 1

n = 2

Now;

= 0.8 – (-0.76)

= 1.56v

Alternative Solution 2

From

= 1.47v

= 1.47v

d) Solution

Alternative Solution 1

= 0.8 – (0.77)

= 0.03v

Alternative Solution 2

Half reactions

n = 1

Now using

E_cell = 0.0122v

c) Solution

= -0.14 – (-2.56)

= 2.42v

Half reactions

2Al + 3Sn²⁺ → 2Al³⁺ + 3Sn

n = 6

Now;

E_cell = 2.418v

EQUILIBRIUM CONSTANT OF GALVANIC CELL

Consider what happens to a Daniel’s cell if we use it to do some electrical work i.e. it can be connected to a small electric motor. After some time the motor will stop. The cell will run down. When this happens and there is no overall transfer of electricity from one half cell to the other. When there is no overall change taking place in a chemical reaction the equilibrium has been established. At equilibrium, electron density of both electrodes is equal and there is no transfer of electricity between the two half cells.

Applying the Nernst equation to a Daniel’s cell

Since the reaction is at equilibrium, the ratio

K_C = 1.67 x 10³⁷

The large value tells us that, the equilibrium lies almost entirely in favour of copper metal and zinc ions.

Generally, the cell at equilibrium at standard temperature is given by the following expression;

Example

Calculate K_C for the following voltaic cell

Given

Cell reaction

From

= -0.12 – (-0.46)

= 0.34v

Now,

K_C = 3.20 x 10¹³

Example

What concentration of will emf of the cell be zero at if concentration of is

3. MEASUREMENT OF pH OF A SOLUTION USING STANDARD ELECTRODE POTENTIAL

pH is the degree of alkalinity, or acidity of a substance. It is obtained by using the hydrogen ion concentration pH = –

The pH of a solution is determined by using any electrode provided its standard electrode potential is known and the concentration of ions in that electrode should be 1M. This will be one of the half cells another half cell is made up hydrogen electrode dipped into the solution whose pH is to be determined.

Reaction:

H₂(g) + 2H⁺(aq) + 2e^– E⁰ = 0.00v

2Ag⁺(aq) + 2e^– → Ag(s) E = 0.8v

H₂(g) + 2Ag⁺(aq) → 2H⁺(aq) + Ag(s) E⁰ = 0.8V

Applying Nernst equation at standard temperature

E_cell = E_cell⁰ – (RT/nF) ln Q

= 0.8 –

Question: Calculate the of the following cell and hydrogen ion concentration.

Zn/Zn²⁺//H⁺/H₂, pt

Zn²⁺/Zn = -0.76v

E_cell = 0.115v

Solution

= 0 – (-0.760)

= 0.76v

= –

Zn(s) → Zn²⁺(aq) + 2e^–

= + x 2

0.115 = 0.76 + 0.0591log [H⁺]^-2

E_cell = 0.8 + 0.0591 x 2(-log[H⁺])

pH= -log[H⁺]

Question 1

Calculate of the following cell-

Zn/Zn²⁺//H⁺/H₂, pt

Zn²⁺/Zn = -0.76v

E_cell = 0.115v

Question 2

If E for Zn /Zn²⁺//Cn²⁺/Cn is 1.1v

1. Calculate the E_cell when concentration of Zn²⁺ is 2M and Cu²⁺ is 0.5M
2. What is E_cell when concentration of Zn is 0.1M, Cu²⁺ is 0.1M

Question 3

Write down the expression for the cell emf for the following reaction;

Briefly explain why the oxidizing power of permanganate (VII) ion is quite sensitive to the concentration of H⁺ in the solution.

ELECTROCHEMICAL SERIES

Is the series of standard electrode potential with respect to their elements from more negative standard electrode potential to the more positive standard electrode potential.

Is the arrangement of electrodes of elements in order of reducing power.

Uses

It is a good guide for predicting reaction that takes place in solution especially displacement reaction.

Displacement reaction

Is a type of reaction in which an atom or element displaces another element or atom in a compound.

Example

What will happen when magnesium ribbon is added to a solution of AgNO₃?

= -2.37v

= 0.8v

The more negative the electrode potential the greater is the reducing power of that element i.e. the more likely it is to give out the electrons and acts as a reducing agent therefore Mg will reduce Ag⁺ ions to Ag(s).

Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s)

Mg(s)/Mg²⁺//Ag⁺(aq)/Ag(s)

= 0.8 – (-2.37)

= 3.17v

Thus, the element higher in electrochemical series will displace the one lower in the series.

2. Displacement of hydrogen from mineral acids

Metals which are higher in the electrochemical series than hydrogen react with acids and replace hydrogen but metals below hydrogen have no action with mineral acids.

Complete the following reaction:-

i. Zn(s) + 2HCl(aq) → ZnCl₂(s) + H₂(g)

ii. Cu(s) + HCl(aq) → No reaction

3. The knowledge of electrochemical series helps in choosing method for extraction of metals.

Higher most metal can’t be extracted from the oxides by chemical reduction process. This is because they are strong reducing agent hence they can be reduced easily from their oxide by electrolysis.

Questions

For electrolysis, fused or molten metal should be used and not aqueous solution. Why?

Answer:-

In aqueous solution, there are H⁺ ions. Hence metals prefer to react with element in lower electrochemical series than the metal itself, hence for electrolysis we use fused or molten metal and not aqueous solution.

CORROSION AND ITS PREVENTION

Corrosion is the deterioration of the metals due to the chemical reactions taking place on the surface. Usually, the process is due to the loss of metal to a solution in some forms by a redox reaction (unwanted redox reactions).

For corrosion to occur on the surface of a metal there must be anodic area where a metal can be oxidized to metal ions as electrons are produced.

Anode area:-

M(s) → Mⁿ⁺ + ne^–

And cathode area where electrons are consumed by any of all of several half reactions.

Cathodic reaction:-

2H⁺(aq) + 2e^– → H₂(g)

2H₂O(l) + 2e^– → 2OH^–(aq) + H₂(g)

4e^– + O₂(g) + H₂O(l) → 4OH^–(aq)

Anodic reactions occur at cracks or around the area with some impurities.

RUSTING OF IRON

The most common corrosion process is rusting of iron.

Rust is hydrated iron (III) Oxide (Fe₂O₃.XH₂O) which appears as a reddish brown substance on the surface of the iron bar.

Both water and air (oxygen) are required for rusting to occur.

The presence of dissolved salts and acid in water increases its conductivity and speeds up the process of rusting.

If the iron object has free access to oxygen and H₂O as in flowing water, a reddish brown ion (III) oxide will be formed which is rust.

For iron:

Anode: Fe(s) → Fe²⁺(aq) + 2e^–

Cathode: O₂(g) + 2H₂O(l) + 4e^– → 4OH^–(aq)

2Fe(s) + O₂(g) + 2H₂O(l) → 2Fe²⁺(aq) + 4OH^–(aq) (rust)

4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) → Fe₂O₃.XH₂O

If oxygen is not freely available: The further oxidation of iron (II) hydroxide is limited to formation of magnetic iron oxide.

6Fe(OH)₂(s) + O₂(g) → 2Fe₃O₄.H₂O + 4H₂O(l)

Fe₃O₄ Black magnetite

Prevention of rusting

Corrosion can’t be made non-spontaneous but it can be prevented by making the rate of the reaction negligible. This can be done by covering the metal surface with protective coating or by providing alternative redox pathways (oxidation-reduction pathways).

Protective coatings are usually of 3 types:

1. Painting: – is the simplest and most common method where a metal surface is properly cleaned and then applied with several layers of rust-proofing paint.
2. Corrosion inhibitors: – these interfere with flow of charges needed for corrosion to take place i.e. phosphate (II) coating on a surface of iron or steel. Using phosphoric acid serves that purpose.
3. Galvanization: (sacrificial protection) a metal can also be protected by coating with a thin film of second metal where the second metal is oxidised instead of the 1^st metal.

Often iron is coated with another metal like zinc, tin or chromium for protection on the surface.

Note: zinc is preferred to tin because zinc protects iron against rusting when its coating has broken down. This is because it has a more negative reduction potential than iron: it acts as a cathode hence it is not changed. This is called cathodic protection.

Tin protects iron only as long as coating is intact. Once the coating is broken down, tin actually promotes corrosion of iron as iron has more negative reduction potential than tin. Thus iron acts as anode and dissolves while tin acts as cathode and does not change.

Factors which affect corrosion:

1. Position of metal in the electrochemical series (E.C.S). The reactivity of metal depends upon its position in the electrochemical series. More the reactivity, the more likely it is to be corroded.
2. Presence of impurities in the metal. The impurities help to set up the voltaic cells which increase the speed of corrosion.
3. Presence of electrolytes: Presence of electrolytes in water also increase the rate of corrosion. E.g. Corrosion of iron in sea water takes place to a larger extent than in distilled water.
4. Presence of CO₂ in water: water containing CO₂ acts as an electrolyte and increases the flow of electrons from one place to another. (CO₂ + H₂O) form carbonic acid which dissolves into ions and hence acts as an electrolyte).

Examples

1. Why do you think zinc on iron is sometimes called sacrificial anode?
2. Explain why blocks of Mg can be attached to walls of ship or iron pipes with the aim of preventing rusting.
3. Tin cans are made of Iron coated with thin film of tin. After a crack occurs in the film, a can corrodes much more rapidly than Zinc coated with Iron. Explain this behavior.
4. Why is it that with enough time, corrosion will always defeat the protection applied to iron?

ANSWERS

Zinc or Iron is sometimes called sacrificial anode because after it wears off, the metal can get exposed and hence starts to undergo rust.

Conductivity in solutions

Electrolytes

These are substances which allow electricity to pass through them in their molten state or in form of their aqueous solution, and undergo chemical decomposition e.g., acids, bases and salts.

Classification of electrolytes

All electrolytes do not ionize by the same extent in the solution. According to this we have strong and weak electrolytes.

Strong electrolytes are those that ionize completely into ions in the solution e.g. Salts, mineral acids, some bases.

Weak electrolytes are those that ionize partially into ions in the solution e.g. organic acids, HCN, Na₄OH.

Electrolytic conduction

When a voltage is applied to the electrodes dipped into an electrolytic solution, ions of the electrolyte move towards their respective electrodes and therefore electric current flows through the electrolytic cell. The process of the electrolyte to conduct electric current is termed as conductance or conductivity.

Like metallic conductors, electrolytic solution also obey Ohm’s law which states that “the strength of the current flowing through a conductor is directly proportional to the potential difference applied across the conductor and inversely proportional to the resistance of the conductor” i.e. V = RI.

Resistance of any conductor is directly proportional to the length L and inversely proportional to the area of cross-section.

R = ρ x (L/A)

ρ = Resistivity

Resistivity is the resistance of a conductor having unit length and unit area of cross-section. SI unit Ohm-meter (Ωm).

Conductance is the measure of the ease with which the current flows through a conductor (λ or Λ).

Conductance is the reciprocal of electric resistance. (G = 1/R)

From the above expression high resistance means low conductance and vice versa.

Conductivity is the reciprocal of resistivity and is also called specific conductance (κ).

R = ρ x (L/A) = (1/κ) x (L/A)

For an electrolytic cell, l is the distance apart between the two electrodes and A is the total area of cross-section of the two electrodes. Therefore, for a given cell, l and A are constant. If the dimensions of the cell are not altered, the ratio (l/A) is referred as cell constant (K).

R = ρ x K

Molar conductance (λm)

Is the conductivity of volume of a solution which contains 1 mole of the solute.

Or

It is the conducting power of the ions produced by dissolving 1 mole of an electrolyte in a solution.

Molar conductance is given by λm = κ x V where V = volume containing 1 mole of a solute and is called dilution.

Concentration of an electrolyte depends on the volume i.e.

C = 1/V

Where C is the concentration in mol/dm³ and V is volume in dm³.

Example

1. What is the dilution of 0.2M NaOH solution?

Solution:

Given λ = 0.2 mol/dm³

V = 1/0.2 = 5 dm³/mol

2. 0.0055M silver nitrate has a molar conductivity of 2.98 x 10^-3 S m² Ω^-1 mol^-1. Calculate conductivity of that solution.

Solution:

κ = λ x c

= 2.98 x 10^-3 x 0.0055

= 1.419 x 10^-4 Ω^-1 cm^-1

ii) Calculate the molar conductivity of 0.3M KOH solution which has a conductivity of 391 Ω^-1 m^-1.

Solution:

λm = κ / c

= 391 / 0.3

= 1303 S m² mol^-1

VARIATION OF MOLAR CONDUCTIVITY WITH CONCENTRATION

The intensity of electricity that can pass through the solution depends on

1. The number of concentration of free ions present in the solution.
2. Speed with which ions move to their respective electrodes.

An increase in concentration gives an increase in total number of solute particles in a given volume of solution and this might well be expected to give an increased conductivity.

Conduction in strong electrolytes.

The molar conductivity is high since strong electrolytes ionize completely into free ions and the molar conductivity increases slightly in dilution. WHY?

In strong electrolytes, there are vast numbers of ions which are close to each other. These ions tend to interfere with each other as they move towards their respective electrodes. The positive ions are held back by the negative ions and vice versa which in turn interrupts their movement to the electrodes (reduce the speed with which they move).

Dilution ions get separated from each other and at an average distance they can move freely or easily.

NOTE:-

As the dilution increases, there comes a time when amount of interference becomes small so that further dilution has no effect.

At this point the molar conductivity remains constant and it is known as molar conductivity at zero concentration λ_∞.

Conduction in weak electrolyte

The molar conductivity is less because there is less number of particles as the minority of particles are dissociated into ions. On dilution the molar conductivity increases as the molecules dissociate more into ions which increases the number of free ions. Therefore for weak electrolytes the molar conductivity depends on the degree of dissociation of molecules into ions.

Question 1

At infinity dilution, will the molar conductivity of strong and weak electrolyte of same concentration be the same?

Answer

NO, the molar conductivity will be different because it also depends on the size of the ions which would either increase or decrease the speed of ions.

Question 2

Why at infinity dilution, the molar conductivity of weak electrolyte remains constant?

Answer

Because at infinity dilution the molecule must have to dissociate into free ions.

MOLAR CONDUCTIVITY AND DEGREE OF DISSOCIATION

The molar conductivity of weak electrolyte is proportional to the degree of dissociation.

i.e. λ = α λ_∞

At infinity dilution, the weak electrolyte is completely ionized.

α = 1 or 100%

Taking ratio of equation (i) and (ii)

Ostwald’s dilution law

It states that “For a weak electrolyte the degree of dissociation is proportional to the square root of reciprocal of concentration”.

Consider a weak binary electrolyte AB (i.e. ethanoic acid) in solution with concentration C (mol/dm³).

AB ⇌ A⁺(aq) + B^–(aq)

1 mole 0 0 start

At equilibrium

K_a = C α² / (1 – α)

For a weak electrolyte, degree of dissociation is very small thus the expression 1 – α ≈ 1

K_a = C α²

Example

1. At 25°C the solution of 0.1M of ethanoic acid has a conductivity of 5.0791 x 10^-2 Sm² mol^-1. Calculate the pH of the solution and dissociation constant K_a (Answer K_a=1.69 x 10^-2 M)

2. A 0.001 mol/dm³ solution of ethanoic acid was found to have molar conductivity of 14.35 Sm² mol^-1. Use this value together with molar conductivity at infinity dilution of ethanoic acid λ_∞ (C₂H₃COOH) is 390.7 Sm² to calculate :

1. Calculate degree of dissociation of acid
2. Equilibrium constant

3. The resistivity of 1M KCl solution is 361Ω and conductivity cell containing such a solution was found to have a resistance of 550Ω

1. Calculate the cell constant
2. The same cell filled with 0.1M ZnSO₄ solution had resistance of 72Ω. What is the conductivity of this solution?

KOHLRAUSCH’S LAW OF INDEPENDENT IONIC MOBILITY

Kohlrausch notes that the difference between molar conductivity at infinity dilution λ_∞ values for the two salts which were strong electrolytes and of the same cation and anion was always constant.

Using the λ_∞ values in D.. cm² mol^-1

This observation shows that each type of ion (cation or anion) contribute a definite amount of molar conductivity of an electrolyte at infinity dilution independently from other ions present in the solution i.e. a fraction of current that an ion carries is always constant and it doesn’t depend on the compound in which it is contained.

Hence Kohlrausch’s law

States that: “The molar conductivity of an electrolyte at infinity dilution is equal to the sum of the molar conductivities of the cation and anion”.

OR State that the molar conductivity at infinity dilution of the solution equals the sum of molar conductivity at infinity dilution of its component ions.

i.e. λ_∞(AB) = λ_∞(A⁺) + λ_∞(B^–)

E.g. = λ_∞(Al³⁺) + 3 λ_∞(SO₄^2-)

Application of Kohlrausch’s law

In direct determination of molar conductivity at infinity dilution for weak electrolytes. Thus by using strong electrolytes we can easily calculate the molar conductivities at infinity dilution for weak electrolytes.

E.g.: The λ_∞ of potassium ethanoate (CH₃COOK), hydrochloric acid (HCl) and potassium chloride (KCl)

CH₃COOH + KCl → CH₃COOK + HCl

λ_∞(CH₃COOH) = λ_∞(CH₃COO^–) + λ_∞(H⁺)

REVIEW QUESTIONS

1. Calculate λ_∞(NH₄OH) given that λ_∞ of three strong electrolytes NaCl, NaOH and NH₄Cl in Ω cm² mol^-1 are 126.4, 248.4, 149.8 respectively.
2. The molar conductivity of 0.093 CH₃COOH solution at 298K is 536 x 10^-4 Sm² mol^-1. The molar conductivity at infinity dilution of H⁺ and CH₃COO^– are 3.5 x 10^-2 and 0.41 x 10^-2 Sm² mol^-1. What is the dissociation constant of CH₃COOH?
3. A 0.05M HF solution has a conductivity of 91.81 mol^-1 m^-1 at 298K. At the same temperature λ_∞ of NaF, NaOH and H₂O are 493, 360 and 162 Sm² mol^-1 respectively. Calculate dissociation constant.
4. The λ_∞ of NaI, λ_∞ of CH₃COONa and λ_∞ of CH₃COO₂Mg are 12.69, 9.10 and 18.78 Sm² mol^-1 respectively at 25°C. What is the molar conductivity of MgI₂ at infinity dilution?