   # Calculate E cell at 298 K for a cell involving 5n and Cu and their ions: Sn(s)|Sn 2+ (aq, 0.25 M)||Cu 2+ (aq, 0.10 M)|Cu(s) (a) 0.47 V (b) 0.49 V (c) 0.50 V ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19.5, Problem 1RC
Textbook Problem
1 views

## Calculate Ecell at 298 K for a cell involving 5n and Cu and their ions:Sn(s)|Sn2+(aq, 0.25 M)||Cu2+(aq, 0.10 M)|Cu(s) (a) 0.47 V (b) 0.49 V (c) 0.50 V

Interpretation Introduction

Interpretation: The Ecell at 298K for a cell involving SnandCu and their ions is to be determined from the given option.

Concept introduction:

• The substance that easily be reduced in a reaction is represented as an oxidizing agent. For metal cations, a good oxidizing agent can be determined by the standard reduction potential values.
• Half-cell: In the electrochemical cell, both oxidation and reduction occurs. Oxidation occurs at the anode and reduction occurs at the cathode
• Standard electrode potential of cell is defined as the difference of reduction potential at cathode to the reduction potential at anode.
• For the cathode and anode reactions, the values are taken from the standard reduction potential table. The standard cell potential is calculated by taking the difference between the standard reduction potential values of the cathode and anode.

Ecell0=Ecathode0Eanode0

• Ecell=E00.0257VnlnQwhere,Qisthereactantquotient=[anode][cathode]
• Ecell=ElectrochemicalcellpotentialE0=Standardelectrochemicalcellpotentialn=numberofelectronspassedfromanodetocathode

### Explanation of Solution

Reason for correct option

There are two half-cell reactions are involved

At anode, oxidation takes place

Sn(s)Sn2+(aq)+2e

At cathode, reduction takes place

Cu2+(aq)+2eCu(s)

E is calculated for the above reaction.

Given,

From standard reduction potential table,

Eanode0=ESn2+/Sn0=0.14V

Ecathode0=ECu2+/Cu0=0.337V

[Sn2+]=0.25M[Cu2+]=0.10M

Ecell0 can be calculated by using the given equation,

Ecell0=Ecathode0Eanode0=ECu2+/Cu0ESn2+/Sn0=0.337V(0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
How is chemosynthesis different from photosynthesis?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

What is the life expectancy of a 16-solar-mass star?

Horizons: Exploring the Universe (MindTap Course List)

Review. A 1.00-m-diameter circular mirror focuses the Suns rays onto a circular absorbing plate 2.00 cm in radi...

Physics for Scientists and Engineers, Technology Update (No access codes included)

Fostering a sense of autonomy in a one-year-old includes allowing the child to explore and experiment with her ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List) 