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)Â +Â 2eâˆ’Â Â Ã Zn2+(aq)Â +Â 2eâˆ’Â Â âˆ’âˆ’âˆ’âˆ’Â atÂ anode,E0anode=Â âˆ’0.76VCu2+(aq)Â +Â 2eâˆ’Â Â Ã Cu(s)Â Â Â Â Â Â âˆ’âˆ’âˆ’âˆ’Â atÂ cathode,E0cathode=Â 0.34Â V

E0cellÂ =Â E0cathodeÂ âˆ’Â E0anodeÂ

= 0.34-(-0.76) =1.10V

Total cell reaction is:

Zn(s)Â +Â Cu2+(aq)(4

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