Maron and Frutton Chapter XV. Transference and Conductance (M&F)

Word Bank

Demal:

Advise: specific conductance of different concentration of KCl solutions (in demal) can be checked in Maron and Frutton book at page 453.

Problems

1. A potential of 110 volts d-c is applied to the terminals of an electric lamp, and a current of 2 amp

is found to flow through the lamp. (a) What is the resistance of the lamp, and (b) how many calories of heat are dissipated per hour?

(a) The first step in solving this problem is to remind the Ohms law

𝑖𝑅 = 𝐸

Where E is the potencia, i is the current and R is the resistance. Now, applying the Omhs law we

obtain the value of the resistance R.

𝑅 =𝐸

𝑖

𝑅 =110 𝑣𝑜𝑙𝑡𝑠

2 𝑎𝑚𝑝= 55 𝑜ℎ𝑚𝑠

(b) To solve this problem it is necessary to use the electrical work equation

𝑤 = 𝐸𝑄

Where w is the done electrical work and Q is the transferred charged. The charge is calculated as follow

1 ℎ𝑟 ×3600 𝑠𝑒𝑐

1 ℎ𝑟×

2 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑠𝑒𝑐= 7,200 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

Now, applying the equation of electrical work we obtain that

𝑤 = 110 𝑣𝑜𝑙𝑡𝑠 × 7200 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 = 792000 𝑣𝑜𝑙𝑡𝑠 × 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑤 = 79200 𝑗𝑜𝑢𝑙𝑒𝑠

𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠× 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑐𝑎𝑙

4.184 𝑗𝑜𝑢𝑙𝑒𝑠= 189,300 𝑐𝑎𝑙

2. A direct current of 0.5 amp flows through a circuit for 10 min. under an applied potential of 30

int. volts. Find the quantity of electricity transported by the current in (a) int. coulombs, (b) abs. coulombs, (c) emu and (d) esu.

(a)

0.5𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑠𝑒𝑐× 10 𝑚𝑖𝑛 ×

60 𝑠𝑒𝑐

1 𝑚𝑖𝑛= 300 𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

(b)

300 𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏

1,000165 𝑖𝑛𝑡.𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠= 299.95 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

(c)

299.95 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 × 10−1𝑒𝑚𝑢

1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 29.995 𝑒𝑚𝑢

(d)

299.95 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×2.9978 × 109𝑒𝑠𝑢

1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 8.9919 × 1011 𝑒𝑠𝑢

3. What is the rate of the dissipation of the energy by the current mentioned in the preceding problem in (a) int. watts, (b) abs. watts, (c) erg sec-1, (d) cal sec-1?

(a)

A watt is a work performed at the rate of 1 joule per second, and is obviously a unit of electrical power.

𝑤 = 𝐸𝑄

If i.t=Q

𝑤 = 𝐸𝑖𝑡

𝑑𝑤

𝑑𝑡= 𝐸𝑖 =

𝐸𝑄

𝑡= 𝑝

𝑝(𝑖𝑛𝑡. 𝑤𝑎𝑡𝑡𝑠) =30 𝑖𝑛𝑡. 𝑣𝑜𝑙𝑡𝑠 × 300 𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

3600 𝑠𝑒𝑐= 2.5 𝑖𝑛𝑡. 𝑤𝑎𝑡𝑡𝑠

(b)

2.5 𝑖𝑛𝑡. 𝑤𝑎𝑡𝑡𝑠 ×1 𝑎𝑏𝑠. 𝑤𝑎𝑡𝑡

0,999835 𝑖𝑛𝑡.𝑤𝑎𝑡𝑡𝑠= 2.5004 𝑎𝑏𝑠. 𝑤𝑎𝑡𝑡𝑠

(c)

2.5 𝑖𝑛𝑡.𝑤𝑎𝑡𝑡𝑠 = 2.5𝑖𝑛𝑡.𝑣𝑜𝑙𝑡𝑠 × 𝑖𝑛𝑡.𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑠𝑒𝑐= 2.5

𝑗𝑜𝑢𝑙𝑒𝑠 × 𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 × 𝑠𝑒𝑐= 2.5

𝑗𝑜𝑢𝑙𝑒𝑠

𝑠𝑒𝑐

2.5𝑗𝑜𝑢𝑙𝑒𝑠

𝑠𝑒𝑐×

1 × 107𝑒𝑟𝑔𝑠

1𝑗𝑜𝑢𝑙𝑒= 2.5 × 107

𝑒𝑟𝑔𝑠

𝑠𝑒𝑐

(d)

2.5𝑗𝑜𝑢𝑙𝑒𝑠

𝑠𝑒𝑐×

1 𝑐𝑎𝑙

4.184 𝑗𝑜𝑢𝑙𝑒𝑠= 0,598

𝑐𝑎𝑙

𝑠𝑒𝑐

4. The value of the faraday is 96,485 abs. coulombs per equivalent, while Abogadro’s number is

6.0235 × 1023. Find the value of the electronic charge in (a) abs. coulombs, (b) int. coulombs, (c) emu, and (d) esu.

(a)

96,486 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

1 𝑚𝑜𝑙𝑒 𝑒−×

1 𝑚𝑜𝑙𝑒 𝑒−

6.0235 × 1023 𝑒−= 1.6018 × 10−19 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

(b)

1.6018 × 10−19 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1.000165 𝑖𝑛𝑡.𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 1.6020 × 10−19 𝑖𝑛𝑡. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

(c)

1.6018 × 10−19 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 × 10−1 𝑒𝑚𝑢

1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 1.6018 × 10−20 𝑒𝑚𝑢

(d)

1.6018 × 10−19 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×2.9978 × 109 𝑒𝑠𝑢

1 𝑎𝑏𝑠. 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 4.8019 × 10−10 𝑒𝑠𝑢

5. A constant direct current flows through an iodine coulometer for a period of 2 hr. At the end of

this time 25.2 cc of 0.08 molar Na2S2O3 are required to react with the liberated I2. What was the current passing through the coulometer?

Inside the coulometer occurs the electrochemical reduction of iodide ions into iodine as it is indicated in the following equation

2𝐼− → 𝐼2 + 2𝑒−

The titration of iodine with sodium thiosulfate follows the equation

2𝑁𝑎2𝑆2𝑂3 + 𝐼2 → 𝑁𝑎2𝑆4𝑂6 + 2𝑁𝑎𝐼

As it is known that the titration has consumed 25.2 cc of 0.08 molar Na2S2O3, we can find the amount of I2 formed during the electrolysis as follow

25.2 𝑐𝑐 ×0.08 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎2𝑆2𝑂3

1000 𝑐𝑐×

1 𝑚𝑜𝑙𝑒 𝐼22 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎2𝑆2𝑂3

×2 𝑚𝑜𝑙𝑒𝑠 𝐼−

1 𝑚𝑜𝑙𝑒 𝐼2= 2.016 × 10−3𝑚𝑜𝑙𝑒𝑠 𝐼−

The first equation says that for 2 moles of iodide ions are produced 2 moles of electrons, and applying the Faraday’s law of the electrolysis we obtain that

2.016 × 10−3 𝑚𝑜𝑙𝑒𝑠 𝐼− ×2 𝑚𝑜𝑙𝑒𝑠 𝑒−

2 𝑚𝑜𝑙𝑒𝑠 𝐼−×

96,845 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑚𝑜𝑙𝑒 𝑒−= 195.24 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

As it is known, the current is i=Qt-1

195.24 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

2 ℎ𝑟×

1 ℎ𝑟

3600 𝑠𝑒𝑐= 0.027 𝑎𝑚𝑝

6. The platinum crucible used in a silver coulometer gains 0.500 g in a certain electrolysis. What

would be the gain in weight of a copper cathode in a cell filled with potassium cuprocyanide [KCu(CN)2] placed in the same circuit.

Firstly, we need to know the amount of charged that have been transferred during the electrolysis process. These value can be obtained as follow

𝐴𝑔+ + 𝑒− → 𝐴𝑔

𝐶𝑢+ + 𝑒− → 𝐶𝑢

0.5 𝑔 𝐴𝑔 ×1 𝑚𝑜𝑙𝑒 𝐴𝑔

107.8683 𝑔 𝐴𝑔×

1 𝑚𝑜𝑙𝑒 𝐶𝑢

1 𝑚𝑜𝑙𝑒 𝐴𝑔×

63.536 𝑔 𝐶𝑢

1 𝑚𝑜𝑙𝑒 𝐶𝑢 = 0.29 𝑔 𝐶𝑢

The weight gain by the cathode copper is 0.29 grams.

7. What volume of O2 would be liberated from an aqueous solution of NaOH by a current of 2 amp flowing for 1 ½ hr? The temperature is 27°C and the total pressure is 1 atm.

The following reaction happens at the anode

4𝑂𝐻− → 𝑂2 + 2𝐻2𝑂 + 4𝑒−

The total charged passed through the circuit can be calculated with the current and the time of the process

2 𝑎𝑚𝑝 = 2𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑠𝑒𝑐× 1.5 ℎ𝑟

3600 𝑠𝑒𝑐

1 ℎ𝑟= 10,800 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

With the total charged in coulombs it is possible to calculate the amount of moles of O2 liberated as

follow

10,800 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠×

1 𝑚𝑜𝑙𝑒 𝑂24 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

= 0.02798 𝑚𝑜𝑙𝑒 𝑂2

Applying the real gas equation PV=nRT we calculate the volume of O2 liberated

𝑉 =0.02798 𝑚𝑜𝑙𝑒𝑠 𝑂2 × 0.082

𝑎𝑡𝑚 ∙ 𝐿𝑚𝑜𝑙 ∙ 𝐾

× 300.15𝐾

1 𝑎𝑡𝑚×

1000 𝑐𝑐

1 𝐿= 688.7 𝑐𝑐

8. (a) How long would it take a current of 1 amp to reduce completely 80 cc of 0.1 molar Fe2(SO4)3?

(b) How many cc of 0.1 molar Kr2CrO7 could be reduced to chromic sulfate. Cr2(SO4)3. By the same quantity of electricity?

(a) The problem asks to completely reduce 80 cc of 0.1 molar of Fe2(SO4)3, it means, to pass the Fe3+

to Fe(s). The following equation depict the process

𝐹𝑒3+ + 3𝑒 − → 𝐹𝑒

So, it is needed three electrons per mole of Fe3+ to be completely reduced. As we know the current

flowing through the circuit is 1 amp (or 1 coulombs per sec), we can determine the time required to

the process to take place knowing the amount of electricity required to completely reduce all the Fe3+.

The amount of electricity required in the process can be calculated using the Faraday’s law of the electrolysis.

80 𝑐𝑐 ×0.1 𝑚𝑜𝑙𝑒 𝐹𝑒3+

1000 𝑐𝑐×

3 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

1 𝑚𝑜𝑙𝑒 𝐹𝑒3+ ×

96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡= 2,315.64 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

Now, we can calculate the time

2,315.64 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑠𝑒𝑐

1 𝑐𝑜𝑢𝑙𝑜𝑚𝑏= 2,315.64 𝑠𝑒𝑐

(b) The process follows the equation below

𝐶𝑟2𝑂72− + 14𝐻3𝑂

+ + 6𝑒 − → 2𝐶𝑟3+ + 21𝐻2𝑂

It can be realized that the Cr passes from the oxidation number 6+ to 3+, so it gains three electrons.

As in the process it is involved to Cr atoms, the total electrons transfer per mole of K2Cr2O7 are six.

Now, applying the Faraday’s law of electrolysis we obtain the volume of 0.1 molar K2Cr2O7 that reduces the same amount of electricity.

2,315.64 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠×

1 𝑚𝑜𝑙𝑒 𝐶𝑟2𝑂72−

6 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠×

1000 𝑐𝑐

0.1 𝑚𝑜𝑙𝑒 𝐶𝑟2𝑂72− = 4 𝑐𝑐

9. What quantity of electricity would be required to reduce 10 g of nitrobenzene completely to aniline?

If the potential drop across the cell is 2 volts, how much energy, in calories, is consumed in the process?4

[Reaction in chemsketch]

The electrochemical reduction of nitrobenzene to aniline required the transference of six electrons.

The first step is to identify how much electricity is needed to the process to take place, and then, it is possible to calculate the electrical work as follow

10 𝑔 𝐶6𝐻5𝑂2𝑁 ×1 𝑚𝑜𝑙𝑒 𝐶6𝐻5𝑂2𝑁

123 𝑔 𝐶6𝐻5𝑂2𝑁×

6 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

1 𝑚𝑜𝑙𝑒 𝐶6𝐻5𝑂2𝑁×

96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡= 46,312.8 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

Now, with the following equation the electrical work is calculated

𝑊 = 𝐸𝑖𝑡 = 𝐸𝑄

𝑊 = 46,312.8 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 × 2𝑗𝑜𝑢𝑙𝑒𝑠

𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠= 92,625.6 𝑗𝑜𝑢𝑙𝑒𝑠

92,625.6 𝑗𝑜𝑢𝑙𝑒𝑠 ×1 𝑐𝑎𝑙

4.184 𝑗𝑜𝑢𝑙𝑒𝑠= 22,138 𝑐𝑎𝑙

10. A 4 molal solution of FeCl3 is electrolyzed between platinum electrodes. After the electrolysis the

cathode portion, weighing 30 g, is 3.15 molal in FeCl3 and 1.00 molal in FeCl2. What are the transport numbers of Fe3+ and Cl- ions?

The problem says that the total weight of the solution inside the cathode is 30 g, so we have that

𝑊 𝐻2𝑂 + 𝑊 𝐹𝑒𝐶𝑙3 + 𝑊 𝐹𝑒𝐶𝑙3 = 30 𝑔

As it is known, the units molal concentration are moles of solute per Kg of solvent. For 1,000g of water we have that

1,000 𝑔 𝐻2𝑂 ×3.15 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒𝐶𝑙3

1000 𝑔 𝐻2𝑂 ×

162.204 𝑔 𝐹𝑒𝐶𝑙31 𝑚𝑜𝑙𝑒 𝐹𝑒𝐶𝑙3

= 517.24 𝑔 𝐹𝑒𝐶𝑙3

1,000 𝑔 𝐻2𝑂 ×1 𝑚𝑜𝑙𝑒 𝐹𝑒𝐶𝑙21,000 𝑔 𝐻2𝑂

×126.751 𝑔 𝐹𝑒𝐶𝑙2

1 𝑚𝑜𝑙𝑒 𝐹𝑒𝐶𝑙2= 126.751 𝑔 𝐹𝑒𝐶𝑙2

So

126.751 𝑔 𝐹𝑒𝐶𝑙2 + 517.24 𝑔 𝐹𝑒𝐶𝑙3 + 1000 𝑔 𝐻2𝑂 = 1643.991 𝑔

Now, we can obtain the percentage of each compound in the solution and then to determine the

amount of each one in the 30 g that we have in the cathode solution. Percentages: 7.71% of FeCl2,

31.46% of FeCl3 and 60.83% of water. In the 30 g of the cathode solution we have 2.313 g FeCl2,

9.483 g FeCl3 and 18.249 g of water.

As we supposed that water does not migrate, the amount of water before the electrolysis at the cathode

were 18.249 g. On the other hand, we have that the transference number is given by the current due

to the ion (I- or I+) divided by the total current of the system (I). As I- = z+n+ev- and I+ = z+n+ev+ due

to the condition of electroneutrality (you can follow a detailed explanation in the post of Transference and Transference numbers), the transference number can be expressed as in the following equation

𝑡+ =𝑣+

𝑣+ + 𝑣−

Because of the Hittorf’s rule, as it is explained in the post of Determination of Transference numbers,

if the velocities of cations and anions are different, the loss in concentration of ions in the cathode

and anode due to migration is proportional to the speed of each ion. This parallelism between

concentration loss due to migration and the velocity of the ion responsible for it leads to Hittorf’s rule, which states that

𝐿𝑜𝑠𝑠 𝑖𝑛 𝑐𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑎𝑡 𝑎𝑛𝑜𝑑𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑚𝑖𝑔𝑟𝑎𝑡𝑖𝑜𝑛

𝐿𝑜𝑠𝑠 𝑖𝑛 𝑎𝑛𝑖𝑜𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑎𝑡 𝑐𝑎𝑡ℎ𝑜𝑑𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑚𝑖𝑔𝑟𝑎𝑡𝑖𝑜𝑛=

𝑣+𝑣−

=𝑡+𝑡−

Glasstone(1942) said that “The total decrease in amount of the electrolyte MA in both compartments

of the experimental cell is equal to the number of equivalents deposited on each electrode; if a

coulometer is included in the circuit, the by the Faraday’s laws the same number of equivalents of material, no matter what its nature, will be deposited. It follows, therefore, that

(detailed explanation in the post of Determination of Transference numbers)

𝐿𝑜𝑠𝑠 𝑖𝑛 𝑐𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑎𝑡 𝑎𝑛𝑜𝑑𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑚𝑖𝑔𝑟𝑎𝑡𝑖𝑜𝑛

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑎𝑠𝑠𝑒𝑑= 𝑡+

Amount of cation equivalent lost at anode due to migration is given by the initial amount of FeCl3 at these compartment and the final amount of the same compound.

The final amount of FeCl3 in grams is 9.351 g as it was determined before, and the initial amount is

18.33 𝑔 𝐻2𝑂 ×4 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒𝐶𝑙3

1000 𝑔 𝐻2𝑂×

162.204 𝑔 𝐹𝑒𝐶𝑙31 𝑚𝑜𝑙𝑒 𝐹𝑒𝐶𝑙3

= 11.893 𝑔 𝐹𝑒𝐶𝑙3

The FeCl3 lost at the cathode was 11.893 g – 9.351g = 2.542 g, or 0.0157 moles.

The formation of FeCl2 describe the amount of Fe3+ lost in the electrolysis process

18.33 𝑔 𝐻2𝑂 ×1 𝑚𝑜𝑙𝑒 𝐹𝑒𝐶𝑙21000 𝑔 𝐻2𝑂

= 0.01833

So, 2,63 x 10-3 moles of Fe3+ have arrived at the cathode due to migration or, what is the same, it is the equivalents lost at the anode due to migration. Now, the transference number is given as follow

𝑡+ =2.63 × 10−3𝑚𝑜𝑙𝑒𝑠 𝐹𝑒3+

0.01833 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡×

3 𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

1 𝑚𝑜𝑙𝑒𝑠 𝐹𝑒3+= 0.45

t- = 1 – 0.45 = 0.55

11. A AgNO3 solution containing 0.00739 g of AgNO3 per gram of H2O is electrolyzed between silver

electrodes. During the electrolysis 0.078 g of Ag plate out at the cathode. At the end of the experiment

the anode portion contains 23.14 g of H2O and 0.236 g of AgNO3. What are the transport number of Ag+ and NO3

- ions.

As in the exercise 10., the transport number is calculated with the formula

𝑡+ =𝑙𝑜𝑠𝑠 𝑐𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑚𝑖𝑔𝑟𝑎𝑡𝑖𝑜𝑛

𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑎𝑠𝑠𝑒𝑑

The total equivalents of current passed are given by the amount of Ag plated out at the cathode, as

follow

0.078 𝑔 𝐴𝑔 ×1 𝑚𝑜𝑙𝑒 𝐴𝑔

107.8683 𝑔 𝐴𝑔×

1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝐴𝑔

1 𝑚𝑜𝑙𝑒 𝐴𝑔= 7.231 × 10−4𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

Now, it is necessary to calculate the equivalents of Ag+ lost at the anode due to migration

As it is supposed that water does not migrate from one part of the cell to other, the total amount of

water at the anode during the experiment has been 23.14 g. It is known the concentration of the initial

solution (0.00739g of AgNO3 per g of water), so it is easy to determine the initial amount of salt at the anode.

23.14 𝑔 𝐻2𝑂 ×0.00739 𝑔 𝐴𝑔𝑁𝑂3

1 𝑔 𝐻2𝑂= 0.171 𝑔 𝐴𝑔𝑁𝑂3

The moles of AgNO3 after the electrolysis in the anode compartment is the moles of initial Ag+ ions

plus the Ag+ ions that dissolved from the anode to the solution after the oxidation process, less the moles of Ag+ that migrated to the cathode.

0.236 𝑔 𝐴𝑔𝑁𝑂3 ×1 𝑚𝑜𝑙𝑒 𝐴𝑔𝑁𝑂3

169.8683 𝑔 𝐴𝑔𝑁𝑂3= 1.389 × 10−3 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂3

Equivalents Ag+ migrated (Eq m)

𝐸𝑞 𝑚 = 7.231 × 10−4 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂3 + 1.006 × 10−3𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂3 − 1.389

× 10−4𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂3 = 3.401 × 10−4 𝑚𝑜𝑙𝑒𝑠 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔𝑁𝑂3 𝑜𝑟 𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

It is possible now to calculate the transport number of the Ag+ ion and nitrate ions as well

𝑡+ =0.0003041

0.0007231= 0.47

𝑡− = 1 − 0.47 = 0.53

12. The transference numbers of the ions of 1.000N KCl were determined by the moving boundary

method using a solution 0.80 N BaCl2 as the following solution. Using a current of 0.0142 amp the

time required for the boundary to sweep through a volume of 0.1205 cc was 1675 sec. What are the

transport numbers of K+ and Cl- ions.

As it was deduced in the post “Determination of Transference numbers” the transference number of

the cation can be determined by the following equation

𝑡+ =𝐿𝑜𝑠𝑠 𝑜𝑓 𝑐𝑎𝑡𝑖𝑜𝑛 𝑎𝑡 𝑡ℎ𝑒 𝑎𝑛𝑜𝑑𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑚𝑖𝑔𝑟𝑎𝑡𝑖𝑜𝑛

𝑇𝑜𝑡𝑎𝑙 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑝𝑎𝑠𝑠𝑒𝑑

The total equivalents of current passed can be determined using a coulometer in the circuit. If it is

given that the current during the experiment was constant and the value was 0.0142 amp and the

process time was 1675 sec, so, we have that

𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒𝑑 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑒𝑑 = 0.0142𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠

𝑠𝑒𝑐× 1675 𝑠𝑒𝑐 = 23.785 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

And applying the Faraday’s law of the electrolysis

23.785 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 ×1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠= 2.465 × 10−4 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

𝑡+ =0.1205 𝑐𝑐 × 1 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡

1000 𝑐𝑐 × 2.465 × 10−4 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠= 0.49

𝑡− = 1 − 𝑡+ = 1 − 0.49 = 0.51

13. The cathode, center, and anode chambers of an electrode of an electrolytic cell contain each 10

milliequivalents of HCl in aqueous solution. What will be the number of milliequivalents of HCl in

each compartment after the passage of 5 milliequivalents of electricity through the cell? Assume that t+ = 0.2 and that H2 is given of at the cathode and Cl2 at the anode.

After the passage of 5 mequivalent of current, there are loss 5 milliequivalents of protons at the

cathode due to the formation of molecular hydrogen and 5 milliquivalents of chloride at the anode.

The process in both cathode and anode are the following, respectively

2𝐻+ + 2𝑒 − → 𝐻2

2𝐶𝑙− → 𝐶𝑙2 + 2𝑒−

As each compartment has to preserve the electroneutrality it is necessary a migration of ions. These migration is given by the transport number of each ion.

Cation

(0.8) x 5 milliquivalents = 4 milliequivalents of H+ migrate

Anion

(0.2) x 5 milliequivalens = 1 milliequivalent of Cl- migrate

In the graph we can follow the process

Each sign (+ and -) represents 1 milliequivalent in the following graph.

Cat (+)

++++++++++ - - - - - - - - - -

++++++++++ - - - - - - - - - -

++++++++++ - - - - - - - - - -

Anod (-)

++++++++++ - - - - -

++++++++++ - - - - - - - - - -

+++++ - - - - - - - - - -

++++++ - - - - - -

++++++++++ - - - - - - - - - -

+++++++++ - - - - - - - - -

At the end of the process the cathode has 6 milliequivalents of HCl, the anode 9 milliequivalents of

HCl and the center chamber 10 milliequivalents of HCl.

14. The conductivity of a cell filled with 0.01 demal (molal) KCl solution gives at 0°C a resistance

of 11,210 ohms. The distance between the electrodes in the cell is 6 cm. Find (a) the cell constant, and (b) the average cross-sectional area of the electrodes.

𝐿(𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒) =1

𝑅=

1

𝜌(

𝐴

𝑙)

R: resistance in omhs

A: electrode area in cm

l: distance between electrodes

1

𝜌= 𝐿𝑠

ρ: specific resistivity (the resistance of a conductor 1 cm in length and with a cross-sectional area of

1sq cm)

Ls: specific conductance (1 cm cube (not cc) of material, and it is expressed in mh-os)

𝐾 =𝑙

𝐴

𝐿𝑠𝐾−1 = 𝐿

𝐾 =𝐿𝑠𝐿

= 𝐿𝑠 × 𝑅 = 11,210 𝑜𝑚ℎ𝑠 × 0.00077362 𝑜𝑚ℎ−1 𝑐𝑚−1 = 8.672 𝑐𝑚−1

𝐴 =𝑙

𝐾=

6 𝑐𝑚

8.672 𝑐𝑚−1= 0.692 𝑐𝑚2

Note: the answer in the book says that the area is 1.45 sq cm but these answer is wrong.

15. A conductivity cell filled with 0.10 demal KCl solution gives at 25°C a resistance of 910 omhs.

What will be the resistance when this cell is filled with a solution whose specific conductance at 25°C is 0.00532 mhos?

The first step is to determine the cell constant K by the same way it was determined in exercise 14th.

𝐾 = 0.0128560 𝑜𝑚ℎ−1 𝑐𝑚−1 × 910 𝑜𝑚ℎ𝑠 = 11.69896 𝑐𝑚−1

After that, it is determined the resistance of the new solution with the same equation as follow

𝐾

𝐿𝑠= 𝑅 =

11.69896 𝑐𝑚−1

0.00531 𝑜𝑚ℎ−1 𝑐𝑚−1= 2203.194 𝑜ℎ𝑚𝑠

16. At 25°C a cell filled with 0.01 demal KCl solution gave a resistance of 484.0 ohms. The following data for NaCl solutions were then taken in the same cell at 25°C.

Normality Resistance 0.0005 10,910

0.0010 5,494 0.0020 2,772

0.0050 1,128.9

(a) Calculate Λ for NaCl at each concentration, and (b) evaluate Λo by plotting Λ against √𝐶 and extrapolating to infinite dilution.

Firstly, it is important to know what is equivalent conductance (Λ). The equivalent conductance of

an electrolyte is defined as the conductance of a volume of solution containing one equivalent weight

of dissolved substance when place between two parallel electrodes 1 cm apart, and large enough to contain between them all the solution.

As we know

𝐾 = 𝐿𝑠𝑅 = 484.0 𝑜ℎ𝑚𝑠 × 0.00140877 𝑜ℎ𝑚−1 𝑐𝑚−1 = 0.682 𝑐𝑚−1

There are and specific Ls for each concentration. As we have that

𝐿𝑠 =𝐾

𝑅

N1/2 N R L Ls Λ

0.02236068 0.0005 10,910 9.1659E-05 6.24972E-05 124.994317

0.03162278 0.0010 5,494 0.00018202 0.000124107 124.107026

0.04472136 0.0020 2,772 0.00036075 0.000245975 122.987734

0.07071068 0.0050 1,128.9 0.00088582 0.00060399 120.797945

The graph represents the variation of the equivalent conductance against the squared root of the

concentration in equivalents per liter. (It is possible to download the excel were this table was made, in the additional information section, doc1).

As the equation of the line is y = -86.233x + 126.87, the equivalent conductance at infinite dilution (Λ0) is 126.86 omhs

-1 cm2.

17. At 25°C the equivalent conductances of dilute NaI solutions ar as follow

Molarity Λ

0.0005 125.36 0.0010 124.25

0.0050 121.25 0.0100 119.24

0.0200 116.70 Find Λ0 of NaI at 25°C.

Sq root Nolarity Λ

0.02236068 125.36

0.03162278 124.25

0.07071068 121.25

0.1 119.24

0.14142136 116.70

y = -86,233x + 126,87

120

121

122

123

124

125

126

127

128

-0,01 0 0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08

Λ

sq root (N)

Equivalent Conductance

y = -72,016x + 126,63

114

116

118

120

122

124

126

128

-0,02 0 0,02 0,04 0,06 0,08 0,1 0,12 0,14 0,16

Λ

Sq root (N)

Equivalent Conductance

Λ0 = 126.63 omhs-1 cm2. The exercise can be downloaded in an excel document (Doc1 – Additional

information).

18. From the following equivalent conductances at infinite dilution at 18°C, find Λ0 for NH4OH at 18°C.

Compound Λ0

Ba(OH)2 228.8 BaCl2 120.3

NH4Cl 129.8

Page 456 M&F. Kohlrausch’s law of independent migration of ions: this law states that at infinite

dilution, where dissociation for all electrolytes is complete and where all interionic affects disappear,

each ion can be considered to migrate independently of its co-ions, and thus to contribute its definite

share to the total equivalent conductance of the electrolyte. Further, if this be the case, the

conductance due to a particular ion at infinite dilution should depend only on the nature of the ion,

and not at all on the co-ion with which it is associated. In other words, the law of independent

migration of ions states that Λ0 of any electrolyte is the sum of the equivalent condutances of the ions

composing it, provided, of course, that the solvent and temperature are the same.

Λ0 = 𝑙+0 + 𝑙−

0

Λ0𝑁𝐻4𝑂𝐻 = Λ0𝑁𝐻4𝐶𝑙 − Λ0𝐵𝑎𝐶𝑙2 + Λ0𝐵𝑎(𝑂𝐻)2

Λ0 𝑁𝐻4𝑂𝐻 = 𝑙+0 𝑁𝐻4

+ −1

2𝑙+

0 𝐵𝑎2+ +1

2𝑙+

0 𝐵𝑎2+ + 𝑙−0 𝐶𝑙− − 𝑙−

0 𝐶𝑙− + 𝑙−0 𝑂𝐻−

Λ0𝑁𝐻4𝑂𝐻 = 𝑙+0 𝑁𝐻4

+ + 𝑙−0 𝑂𝐻−

Λ0𝑁𝐻4𝑂𝐻 = 129.8 − 120.3 + 228.8 = 238.8

19. At 25°C the equivalent conductance of a 0.02 molar AgNO3 solution is 128.7, while the transport

number of Ag+ is 0.477. Calculate the equivalent ionic conductance and the ionic mobilities of Ag+ and NO3

- in a 0.02 molar solution of AgNO3.

Λ AgNO3 = 𝑙+𝐴𝑔+ + 𝑙−𝑁𝑂3

−

Λ AgNO3 = 𝑡+Λ AgNO3 + 𝑡−Λ AgNO3

𝑙+𝐴𝑔+ = 0.477 × 128,7 = 61,3899

𝑙−𝑁𝑂3− = 0.523 × 128.7 = 67.3101

20. In measuring the mobility of H+ ions by the moving boundary procedure, it I observed that the

boundary moves a distance of 4 cm in 12.52 min. The voltage drop across the cell before the formation

of the boundary is 16 volts. The distance between electrodes is 9.6 cm. Calculate from these data the mobility and ionic conductance of hydrogen ions.

If the potential drop before the formation of the boundary is 16 volts and the distance between the

electrodes is 9.6 cm, the voltage drop per centimeter (V) is

𝑉 =16 𝑣𝑜𝑙𝑡𝑠

9.6 𝑐𝑚= 1.666 𝑣𝑜𝑙𝑡𝑠 𝑐𝑚−1

Supposing that the protons only moved in the y-axe of the plane (a linear movement), the drift velocity of the ions is given by

𝑢+ =4 𝑐𝑚

12.52 𝑚𝑖𝑛×

1 𝑚𝑖𝑛

60 𝑠𝑒𝑐= 5.3248 × 10−3

𝑐𝑚

𝑠

As the mobility depends of the potential drop, the formula to determine it is the following

𝑣+ =𝑢+𝑉

=5.3248 × 10−3

𝑐𝑚𝑠𝑒𝑐

1.666 𝑣𝑜𝑙𝑡𝑠 𝑐𝑚−1= 0.0032

𝑐𝑚2

𝑣𝑜𝑙𝑡𝑠 𝑠𝑒𝑐

21. At 18°C the mobility at infinite dilution of the ammonium ion is 0.00066 cm / second, while that

of the chlorate ion is 0.00057 cm / second. Calculate Λ0 of ammonium chlorate and the transport numbers of the two ions.

As it is deduced in the post “Absolute velocity of ions (or mobility)” the mobility can be expressed

in terms of equivalent ionic conductance

𝑣+ =𝑙+

∝ 𝐹

𝑣− =𝑙−

∝ 𝐹

The ammonium chloride is a strong electrolyte so in aqueous solution at infinite dilution it is

completely dissociated. In this particular case α = 1. On the other hand, it is known that

Λ = 𝑙+ + 𝑙− = 𝑡+Λ + 𝑡−Λ

So

𝑣+𝐹

Λ= 𝑡+

𝑣−𝐹

Λ= 𝑡−

From those equations it is possible to calculate the value of the equivalent conductance at infinite dilution of the salt and the transport number of each ion.

Λ0𝑁𝐻4(𝑇=18°𝐶) = 𝐹(𝑣+ + 𝑣−) = 96,485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠(0.00066𝑐𝑚2

𝑣𝑜𝑙𝑡𝑠 𝑠𝑒𝑐+ 0.00057

𝑐𝑚2

𝑣𝑜𝑙𝑡𝑠 𝑠𝑒𝑐)

= 118.67𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 𝑐𝑚2

𝑣𝑜𝑙𝑡𝑠 𝑠𝑒𝑐

The transport numbers can be calculated as follow

𝑡+ =0.00066 𝑐𝑚2 𝑣𝑜𝑙𝑡𝑠−1 𝑠𝑒𝑐1 × 96485 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠

118.67𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 𝑐𝑚2

𝑣𝑜𝑙𝑡𝑠 𝑠𝑒𝑐

= 0.537

𝑡− = 1 − 𝑡+ = 0.463

22. For AgNO3 in aqueous solution Λ0 = 133.36 at 25°C. Using the Onsager equation find Λ for a

0.0010 molar solution at the same temperature, and compare the result with the experimentally observed value of Λ = 130.52

(See the post “Interionic Attraction Theory of Conductance” for more details). For the special case of

1-1 electrolytes in water at 25°C, for which z+ = 1 and z- = 1, D = 78.55, and η = 0.008949 poise, the equation of Onsager is reduced to the following expression

Λ = Λ0 − [θΛ0 + 𝜎]√𝐶

θ = 0.2273 and σ = 59.78

Λ = 133.36 − [0.2273 × 133.36 + 59.78]√0.0010 = 130.51

23. At 25°C the resistance of a cell filled with 0.01 demal of KCl is 525 ohms. The resistance of the

same cell filled with 0.1N NH4OH is 2030 ohms. What is the degree of dissociation of NH4OH in this

solution?

NH4OH is a weak electrolyte so

∝ =Λ

Λ𝑒

Ls KCl 0.01N at 25°C = 0.00140877

𝐾 = 𝐿𝑠𝑅 = 0.00140877 × 525 = 0.7396 𝑐𝑚−1

Λ𝑒 = 𝑙+0 + 𝑙−

0 = 73.4 + 198 = 272.3

𝐿𝑠(𝑁𝐻4𝑂𝐻 0.1𝑁) =0.7396 𝑐𝑚−1

2030 𝑜𝑚ℎ𝑠= 3.643 × 10−4

Λ = 𝐿𝑠𝑉𝑒 =1000𝐿𝑠

𝐶=

1000 × 3.643 × 10−4

0.1= 3.643

∝ =Λ

Λ𝑒=

3.643

272.3= 0.0134

24. What will be the resistance of the cell used in the preceding problem when it is filled with H2O

having a specific conductance of 2 x 10-6 mho?

𝑅 =0.7396 𝑐𝑚−1

2 × 10−6 𝑚ℎ𝑜= 369,800

25. The specific conductance at 25°C of a saturated aqueous solution of SrSO4 is 1.482 x 10-4 mho,

while that of the H2O used is 1.5 x 10-6 mho. Using the data in Table 6, determine at 25°C the solubility

in grams per liter of SrSO4 in water.

𝐿𝑠(𝑠𝑎𝑙𝑡) = 𝐿𝑠 − 𝐿𝑠(𝐻2𝑂)

𝐿𝑠(𝑠𝑎𝑙𝑡) = (1.482 × 10−4 − 1.4 × 10−6) 𝑚ℎ𝑜 = 1.467 × 10−4 𝑚ℎ𝑜

The equivalent conductance of the salt is

Λ0 =1

2𝑙+

0 𝑆𝑟2+ +1

2𝑙−

0 𝑆𝑂42− = 58.46 + 79.8 = 138.26

Since the solution is very dilute (because of the difficult solubility of the salt), the equation to determine the concentration is

𝐶 =1000𝐿𝑠

Λ0=

1000 × 1.467 × 10−4

138.26= 1.061 × 10−3

𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

𝐿×

183,68 𝑔𝑟𝑎𝑚𝑠 𝑆𝑟𝑆𝑂42 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠

= 0.097 𝑔𝑟𝑎𝑚𝑠 𝑙𝑖𝑡𝑒𝑟−1

26. In the titration of 25.0 cc (diluted to 300 cc) of NaC2H3O2 solution with 0.0972 N HCl solution the following data were found:

Volume of HCl Used Conductace X 104

10.0 3.32

15.0 3.38

20.0 3.46

45.0 4.64

50.0 5.85

55.0 7.10

What is the strength of the Na2H3O2 solution in moles per liter?

To determine the strength of these salt, the first step is to plot the volume of HCl used during the titration against the conductance

+

Once the graph is done, the next step is to get the linear equation of the data before and after the final

point of the titration. As the intercept point of the two lines is the final point, the next step is to match both equations and to find the x (volume of HCl used) to achieve the final point, as follow

140𝑥 + 31767 = 2460𝑥 − 64367

31767 + 64367 = 2460𝑥 − 140𝑥

96134 = 2320𝑥

𝑥 =96134

2320= 41.44

41.44 cc is the volume of HCl 0.0972 N required to neutralize the solution of sodium acetate. With

this information it is possible to calculate the strength of the sodium acetate before de dilution.

41.44 𝑐𝑐 𝐻𝐶𝑙 ×0.0972 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙

1000 𝑐𝑐×

1 𝑚𝑜𝑙𝑒 𝑁𝑎𝐶2𝐻3𝑂21 𝑚𝑜𝑙𝑒 𝐻𝐶𝑙

= 4.028 × 10−3 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶2𝐻3𝑂2

4.028 × 10−3 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝐶2𝐻3𝑂225 𝑐𝑐

×1000 𝑐𝑐

1 𝑙𝑖𝑡𝑒𝑟= 0.16 𝑀

𝐼 =0.16 𝑀 × 12 + 0.16 𝑀 × (−1)2

2= 0. 16 𝑀