The reason behind the absence of + 1 charge over the given metal for a given amount of metal that is deposited when a given amount of current is passed for a certain period of time is to be stated. Concept introduction: The non-spontaneous reaction takes place in an electrolytic cell in which there occurs conversion of electrical energy into chemical energy and this is used for the electrolysis of a metal. The charge generated in the cell is calculated as, Q = I t When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by, W = Z Q = Z I t The value of Z is given as, Z = Molar mass n × 96 , 485 To determine: The reason behind the absence of + 1 charge over the given metal for a given amount of metal that is deposited when a given amount of current is passed for a certain period of time.

Question

Want to see the full answer?

Answer 1

Question

Chapter 17, Problem 130AE

Interpretation Introduction

Interpretation:

The reason behind the absence of +1 charge over the given metal for a given amount of metal that is deposited when a given amount of current is passed for a certain period of time is to be stated.

Concept introduction:

The non-spontaneous reaction takes place in an electrolytic cell in which there occurs conversion of electrical energy into chemical energy and this is used for the electrolysis of a metal.

The charge generated in the cell is calculated as,

Q=It

When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by,

W=ZQ=ZIt

The value of Z is given as,

Z=Molar massn×96,485

To determine: The reason behind the absence of +1 charge over the given metal for a given amount of metal that is deposited when a given amount of current is passed for a certain period of time.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Part 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answer

Part 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00

Calculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/mole