Solved Examples on Electrolysis

Example 1: How many coulombs are needed for 40.5 g of aluminium to react when the electrode is:

Al³⁺ + 3e^– ⇒ Al

Solution:

1 mol of Al requires 3 mol of electrons or 3 × 96500 C

1 mol of Al = 27g

27g of Al require =3 × 96500 C

40.5g of Al require =(3*96500C × 40.5)/27 = 434,250 C

Example 2: In the electrolysis of acidic water, it is desired to obtain hydrogen at 1cc sec at the STP position. What should be the current pass?

Solution:

2H⁺ + 2e^– ⇒ H₂

1 mol of H₂ or 22400 cc of H₂ at STP requires = 2 × 96500 C

1cc of H₂ at NTP requires = (2×96500)/22400 = 8.616 C

Now, Q = I × t

I = Q/t 

I = 8.616/1

I = 8.616 ampere

Example 3: How many moles of mercury will be produced by galvanic isolation 1.0 M Hg(NO₃)₂ solutions with a current of 2.00 A for 3 hours?

Solution:

Hg²⁺ + 2e^– ⇒ Hg

Charged Passed(Q) = I × t(sec)

Q  = 2.0 A × 3.0 × 60 × 60 = 21600 C

2 × 96500 C of Charge produce = 1 mol of Mercury

21600 C of Charge will produce = 1× 21600 / (2×96500) = 0.112mol of Mercury

Example 4: A solution of CuSO₄ is electrolyzed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Solution:

Current (I) = 1.5 A

Time (t) = 10 min = 10 × 60 = 600s 

amount of electricity passed = I × t = (1.5 A) × (600 s) = 900 C (A s = C)

Copper is Deposited: Cu⁺² + 2e^– → Cu(s)

2 mol of Electrons or 2 × 96500 C of current deposit copper = 63.56 g

900 C of current will deposit copper = 63.56/(2 ×96500) = 0.296 g

Example 5: Calculate how long it will take to deposit 1.0 g of chromium when a current of 1.25 A flows through a solution of chromium (III) sulphate. (Molar mass of Cr=62).

Solution:

Cr³⁺+3e^– → Cr(s) 

3 mol of electricity are needed to deposit 1 mol of Cr,

52 g of Cr require current = 3 × 96500 C

1g of Cr will require current = (3 × 96500)/52  = 5567.3 C

Number of Coulombs = Current × t

Time (s) required = No. of Coulombs / Current 

Time (s) required = 5567.3 C / 1.25 (Ampere)

                            = 4453.8s or 1.24hr

Example 6: How many hours does it take to reduce 3 mol of Fe with 2.0 A current? (F= 96500 C) Solution: Reduction of Fe³⁺ to Fe²⁺.

Solution:

Fe³⁺ + e^– → Fe²⁺ 

Reduction of 1 mol of Fe³⁺ requires = 96500 C

Reduction of 3 mol of Fe³⁺ require = 3 × 96500 C = 2.895 × 10°C

Quantity of electricity = Current × Time

2.895 × 10 = 2 × Time

Time = 2.895 × 10⁵ 

            = 14475 × 10⁵ s

Time = (14475 × 10⁵ )/(60 × 60) 

        = 40.21 hours

Electrolysis is the process of decomposing the ionic compound into its constituent elements by passing the electric current into the solution of the ionic compound. The concept of electrolysis was first given by the famous scientist of the 19th century Michael Faraday. It is a chemical process that uses electrical energy to bring changes in the chemical reaction. Electrolysis is used to separate components of the ionic compounds.

In this article, we will learn about, electrolysis, its process, faraday’s law of electrolysis and others in detail in this article.