   Chapter 13, Problem 13.23PAE Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
ISBN: 9781337398909 Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
ISBN: 9781337398909
Textbook Problem

One half-cell in a voltaic cell is constructed from a copper wire dipped into a 4.8 × 10 − 3 M solution of Cu(NO3)2.The other half-cell consists of a zinc electrode in a 0.40 M solution of Zn(NO3)2. Calculate the cell potential at 298 K.

Interpretation Introduction

Interpretation: Toassign the Electromotive force with the help of Electrochemical series.

Concept Introduction:

• In a galvanic cell: Oxidation occurs at anode and reduction occurs at cathode.
• In a cell representation, the left half represents anodic reaction while the right half represents cathodic reaction.
• Higher is the reduction potential, the electrode acts as a cathode , the process of reduction takes place at that electrode and it acts as an oxidizing agent
• Lower is the reduction potential, the electrode act as an anode, the process of oxidation takes place at that electrode and it acts as reducing agent.
• E0cell = E0cathode  E0anode (When both the values are for reduction potential)
• Nernst equation at 298K is given by:Ecell  = E0cell (0.0591/n) log([product]/[reactant])
Explanation

The half cell reaction is given by:

Zn(s) + 2e  àZn2+(aq) + 2e   at anode,E0anode= 0.76VCu2+(aq) + 2e  àCu(s)       at cathode,E0cathode= 0.34 V

E0cell = E0cathode  E0anode

= 0.34-(-0.76) =1.10V

Total cell reaction is:

Zn(s) + Cu2+(aq)(4

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 