English

A Steady Current of 2 Amperes Was Passed Through Two Electrolytic Cells X and Y Connected in Series Containing Electrolytes Feso4and Znso4 Until 2.8g of Fe Deposited at

Advertisements
Advertisements

Question

A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4and ZnSO4 until 2.8g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. 
(Molar mass: Fe=56g mol-1,Zn=65.3g mol-1,1F=96500C mol-1)

Answer in Brief
Advertisements

Solution

I=2 A
W1= 2.8 g
Fe2+(aq)+2e-→Fe(s)
96500×2 C of charge is required to deposit=56 g of Fe
9650 C of charge is required to deposit=2.8 g of Fe
∵ Q `=1"t"  "or"  "t" 9650/2=4825"s"`
Using Faraday's second law of electrolysis,
`("W"_1("Weight of Fe deposited"))/("W"_2("Weight of Zn deposited")`
`=("E"_1("Equivalent weight of Fe"))/("E"_2("Equivalent weight of Fe")`
`2.8/"W"_2 =(56/5)/((65.3)/2)=56/65.3`
Or W2 = 3.265 g

shaalaa.com
  Is there an error in this question or solution?
2018-2019 (March) 56/1/1

APPEARS IN

RELATED QUESTIONS

Define “Molar conductivity”.


Why conductivity of an electrolyte solution decreases with the decrease in concentration ?


The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Concentration/M 0.001 0.010 0.020 0.050 0.100
102 × κ/S m−1 1.237 11.85 23.15 55.53 106.74

Calculate ∧m for all concentrations and draw a plot between ∧m and c1/2. Find the value of `Lambda_m^0`.


\[\ce{Λ^0_m H2O}\] is equal to:

(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]

(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]

(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]

(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]


Molar conductivity of ionic solution depends on:

(i) temperature.

(ii) distance between electrodes.

(iii) concentration of electrolytes in solution.

(iv) surface area of electrodes.


Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.


Molar conductivity of substance “A” is 5.9 × 103 S/m and “B” is 1 × 10–16 S/m. Which of the two is most likely to be copper metal and why?


The variation of molar conductivity with concentration of an electrolyte (X) m aqueous solution is shown in the given figure.

The electrolyte X is ______.


Assertion (A): Molar conductivity decreases with increase in concentration.

Reason (R): When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.


The following questions are case-based questions. Read the passage carefully and answer the questions that follow:

Rahul set up an experiment to find the resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from the null point, he also calculated the conductivity K and molar conductivity ∧m and recorded his readings in tabular form.
S. No. Conc.
(M)
k S cm−1 m S cm2 mol−1
1. 1.00 111.3 × 10−3 111.3
2. 0.10 12.9 × 10−3 129.0
3. 0.01 1.41 × 10−3 141.0

Answer the following questions:

(a) Why does conductivity decrease with dilution? (1)

(b) If `∧_"m"^0` of KCl is 150.0 S cm2 mol−1, calculate the degree of dissociation of 0.01 M KCI. (1)

(c) If Rahul had used HCl instead of KCl then would you expect the ∧m values to be more or less than those per KCl for a given concentration? Justify. (2)

OR

(c) Amit a classmate of Rahul repeated the same experiment with CH3COOH solution instead of KCl solution. Give one point that would be similar and one that would be different in his observations as compared to Rahul. (2)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×