   # The half-cells Fe 2+ (aq, 0.024 M)|Fe(s) and H + (aq, 0.056 M)|H 2 (1.0 bar) are linked by a salt bridge to create a voltaic cell. Determine the cell potential, E cells at 298 K. ### 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 2CYU
Textbook Problem
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## The half-cells Fe2+(aq, 0.024 M)|Fe(s) and H+(aq, 0.056 M)|H2(1.0 bar) are linked by a salt bridge to create a voltaic cell. Determine the cell potential, Ecells at 298 K.

Interpretation Introduction

Interpretation: The Ecell at 298K for the given cell is to be calculated.

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

There are two half-cell reactions are involved

At anode, oxidation takes place

Fe(s)Fe2+(aq)+2e

At cathode, reduction takes place

2H+(aq)+2eH2(g)

The overall reaction is,

Fe(s)+2H+(aq)Fe2+(aq)+H2(g)

Ecell is calculated for the above reaction.

Given,

From standard reduction potential table,

Eanode0=EFe2+/Fe0=0.44V

Ecathode0=EH+/H0=0.00V

[H+]=0.056M[Fe2+]=0.024MPH2=1

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