   # Use the table of standard reduction potentials (Appendix M) to calculate Δ r G ° for the following reactions at 298 K. (a) 3 Cu(s) I 2 NO 3 − (aq) + 8 H + (aq) → 3 Cu 2+ (aq) + 2 NO(g) + 4 H 2 O( ℓ ) (b) H 2 O 2 (aq) + 2 Cl − (aq) + 2 H+(aq) → Cl 2 (g) + 2 H 2 O( ℓ ) ### 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, Problem 87GQ
Textbook Problem
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## Use the table of standard reduction potentials (Appendix M) to calculate ΔrG° for the following reactions at 298 K. (a) 3 Cu(s) I 2 NO3− (aq) + 8 H+(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(ℓ) (b) H2O2(aq) + 2 Cl− (aq) + 2 H+(aq) → Cl2(g) + 2 H2O(ℓ)

(a)

Interpretation Introduction

Interpretation:

The ΔrG0 for the following reaction has to be determined.

(a) 3Cu(s) + 2NO3-(aq) + 8H+(aq) 3Cu2+(aq) + 2NO(g) + 4H2O(l).

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

### Explanation of Solution

The given chemical reaction is as follows.

3Cu(s) + 2NO3-(aq) + 8H+(aq) 3Cu2+(aq) + 2NO(g) + 4H2O(l)

First, we have to find the oxidized and reduced elements.

Oxidation state:

NO3-x + 3(-2)= -1x = +5

From the given reaction , Cu is being oxidized to Cu2+.NO3- is also being converted to NO and H2O.

Let’s write an each half cell reaction.

At anode:Oxiation : 3Cu2+(aq) + 6e-3Cu (s)At cathode:Reduction : 2NO3 (aq) + 8H++6e-2NO(g) + 4H2O

Let’s calculate the Ecello of the reaction

(b)

Interpretation Introduction

Interpretation:

The ΔrG0 for the following reaction has to be determined.

(b) H2O2(aq) + 2Cl-(aq) + 2H+(aq) Cl2(g) + 2H2O(l).

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

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