   # You have a concentration cell in which the cathode has a silver electrode with 0.10 M Ag + . The anode also has a silver electrode with Ag + ( aq ), 0.050 M S 2 O 3 2− , and 1.0 × 10 −3 M Ag(S 2 O 3 ) 2 3− . You read the voltage to be 0.76 V. a. Calculate the concentration of Ag + at the anode. b. Determine the value of the equilibrium constant for the formation of Ag(S 2 O 3 )2 3− Ag + ( a q ) + 2S O 2 3 2 − ( a q ) ⇌ Ag(S2O 3 ) 3 − 2 ( a q ) K = ? ### 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 84E
Textbook Problem
5 views

## You have a concentration cell in which the cathode has a silver electrode with 0.10 M Ag+. The anode also has a silver electrode with Ag+(aq), 0.050 M S2O32−, and 1.0 × 10−3 M Ag(S2O3)23−. You read the voltage to be 0.76 V.a. Calculate the concentration of Ag+ at the anode.b. Determine the value of the equilibrium constant for the formation of Ag(S2O3)23− Ag + ( a q ) + 2S O 2 3 2 − ( a q ) ⇌ Ag(S2O 3 ) 3 − 2 ( a q )         K = ?

(a)

Interpretation Introduction

Interpretation:

A concentration cell with silver electrode as cathode and anode with different concentrations of different ions is given. Various questions based on the given concentrations and cell potential are to be answered.

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 . The value of equilibrium constant helps to predict the extent of the reaction.

To determine: The concentration of Ag+ at the anode.

### Explanation of Solution

Given

The cell potential is measured as 0.76V .

The reaction taking place on cathode,

Ag+(0.10M)+eAg

The reaction taking place at anode,

AgAg+(aq)+e

Add both the reduction half and oxidation half-reaction.

Ag+(0.10M)+eAgAgAg+(aq)+e

The overall cell reaction is,

Ag+(0.10M)Ag+(aq)

The E°cell is zero, as the electrodes are same on both the sides.

The reaction involves the transfer of 1 mole of electrons and the value of Ecell=0.76V .

The cell potential is calculated using the Nernst equation,

E=E°(0.0591n)log(Q)

Where,

• E is the cell potential.
• E° is the cell potential at the standard conditions.
• Q is the activity of the species in the cell.
• n is the number of electrons involved in the reaction.

Substitute the values of Eo , n and Q in above equation as,

E=E°(0.0591n)log(Q)0

(b)

Interpretation Introduction

Interpretation:

A concentration cell with silver electrode as cathode and anode with different concentrations of different ions is given. Various questions based on the given concentrations and cell potential are to be answered.

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 . The value of equilibrium constant helps to predict the extent of the reaction.

To determine: The value of equilibrium constant for the formation of Ag(S2O3)23 .

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