   # Calculate K sp for iron(II) sulfide given the following data: ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 17, Problem 89E
Textbook Problem
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## Calculate Ksp for iron(II) sulfide given the following data: Interpretation Introduction

Interpretation:

The reduction potential of two half-cell reaction involving iron is given. The value of Ksp for iron sulfide (II) is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

This relation is further used to determine the relation between ΔG° and K , ΔG° and E°cell .

Solubility product is applied only for those ionic compounds that are sparingly soluble. The product of solubility of ions is called solubility product and solubility is present in moles per liter.

To determine: The value of Ksp for iron sulfide (II).

### Explanation of Solution

Consider the given reactions,

FeS(s)+2eFe(s)+S2(aq)E°=1.01V

Fe2+(aq)+2eFe(s)E°=0.44V

The reaction taking place at cathode is,

Fe2+(aq)+2eFe(s)E°red=0.44V

The reaction taking place at anode is,

Fe(s)+S2(aq)FeS(s)+2eE°ox=1.01V

Add both the reduction and oxidation half-reaction to get the final equation,

Fe2+(aq)+S2(aq)FeS(s)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=1.01V+(0.44V)=0.57V

The reaction involves the transfer of 2 moles of electrons transfer

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