Chapter 19, Problem 85GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calculate the cell potential for the following cell:Pt|H2(P = 1 bar)| H+(aq, 1.0 M)||Fe3+(aq, 1.0M), Fe2+(aq, 1.0M)|PtWill this reaction be more or less favorable at lower pH? To determine this, calculate the cell potential for a reaction in which [H+(aq)| is 1.0 × 10−7M.

Interpretation Introduction

Interpretation:

The celll potential for the given cell has to be determined in which [H+(aq)] is1.0×10-7 M and has to be identify whether the reaction is to be more or less favourable at the lower pH.

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

The relation between solubility product Ksp and equilibrium constant is as follows.

Ksp= e+lnK

Explanation

The given voltaic cell is as follows.

PtÂ |Â H2(P=1bar)Â |Â H+(aq,1.0M)Â ||Â Fe3+(aq,Â 1.0Â M),Â Fe2+Â (aq,Â 1.0Â M)Â |Â Pt

From the given voltaic cell , the reduction potential of H+ is less than that of the Fe3+(0.771Â V).

Therefore, PtÂ |Â H2,H+ act as anode and Pt | Fe3+,Fe2+ half cell act as cathode.

Letâ€™s write the half reactions and their reduction potential values:

AtÂ cathode:ReductionÂ :Â 2Fe3++2e-Â â†’2Fe2+;E0=Â 0.771Â VAtÂ anode:Oxidation:Â 2H++2e-Â â†’H2;E0=0.00Â V

Letâ€™s calculate the Ecell0:

Ecell0=Â Ecathodeo-Â Eanodeo=Â 0.771Â VÂ -Â 0.00Â V=Â 0

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