   # You have a concentration cell with Cu electrodes and [Cu 2+ ] = 1.00 M (right side) and 1.0 × 10 −4 M (left side). a. Calculate the potential for this cell at 25°C. b. The Cu 2+ ion reacts with NH 3 to form Cu(NH3) 2+ by the following equation: Cu 2 + ( a q ) + 4 NH 3 ( a q ) ⇌ Cu ( NH 3 ) 4 2 + ( a q ) K = 1.0 × 10 13 Calculate the new cell potential after enough NH 3 is added to the left cell compartment such that at equilibrium [NH 3 ] = 2.0 M . ### 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 148CP
Textbook Problem
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## You have a concentration cell with Cu electrodes and [Cu2+] = 1.00 M (right side) and 1.0 × 10−4M (left side).a. Calculate the potential for this cell at 25°C.b. The Cu2+ ion reacts with NH3 to form Cu(NH3)2+ by the following equation: Cu 2 + ( a q ) + 4 NH 3 ( a q ) ⇌ Cu ( NH 3 ) 4 2 + ( a q ) K = 1.0 × 10 13 Calculate the new cell potential after enough NH3 is added to the left cell compartment such that at equilibrium [NH3] = 2.0 M.

(a)

Interpretation Introduction

Interpretation:

A concentration cell having Copper electrodes at different concentration is given. The cell potential for the given concentration cell at the given concentration of ions and the value of new cell potential after enough NH3 is added to the left compartment at equilibrium condition is to be calculated.

Concept introduction:

A cell consisting of electrodes of same type which have different concentration of ions is called concentration cell. The difference in ion concentration in different compartments is the driving force for the formation of the concentration cell.

To determine: The potential for the given concentration cell.

The value of cell potential for the given cell is 0.118V_ .

### Explanation of Solution

Given,

The temperature at which the cell reaction takes place is 25°C .

The reaction taking place at the cathode is,

Cu2+(1.00M)+2eCuE°red=0.34V

The reaction taking place at the anode is,

CuCu2+(1.0×104M)+2eE°ox=0.34V

Add both the reduction and oxidation half-reactions as,

Cu2+(1.00M)+2eCuCuCu2+(1.0×104M)+2e

The final equation is,

Cu2+(1.00M)Cu2+(1.0×104M)

The value of E°cell is given as,

E°cell=E°ox+E°red

Where,

• E°ox is the oxidation potential of the electrode.
• E°red is the reduction potential of the electrode

(b)

Interpretation Introduction

Interpretation:

A concentration cell having Copper electrodes at different concentration is given. The cell potential for the given concentration cell at the given concentration of ions and the value of new cell potential after enough NH3 is added to the left compartment at equilibrium condition is to be calculated.

Concept introduction:

A cell consisting of electrodes of same type which have different concentration of ions is called concentration cell. The difference in ion concentration in different compartments is the driving force for the formation of the concentration cell.

To determine: The value of new cell potential after enough NH3 is added to the left compartment at equilibrium condition.

The value of new cell potential for the given cell is 0.38V_ .

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