Exercise 18.65 Use tabulated half-cell potentials to calculate G rxn for each of

Question

Exercise 18.65

Use tabulated half-cell potentials to calculate Grxn for each of the following reactions at 25 C.

Part A

Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)

SubmitMy AnswersGive Up

Part B

Br2(l)+2Cl(aq)2Br(aq)+Cl2(g)

Express your answer using two significant figures.

SubmitMy AnswersGive Up

Part C

MnO2(s)+4H+(aq)+Cu(s)Mn2+(aq)+2H2O(l)+Cu2+(aq)

Express your answer using two significant figures.

Exercise 18.65

Use tabulated half-cell potentials to calculate Grxn for each of the following reactions at 25 C.

Part A

Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)

Grxn =   kJ  

SubmitMy AnswersGive Up

Part B

Br2(l)+2Cl(aq)2Br(aq)+Cl2(g)

Express your answer using two significant figures.

Grxn =   kJ  

SubmitMy AnswersGive Up

Part C

MnO2(s)+4H+(aq)+Cu(s)Mn2+(aq)+2H2O(l)+Cu2+(aq)

Express your answer using two significant figures.

Grxn =   kJ  

Explanation / Answer

a)
Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)
dG rxn = -nFE0 cell
dG rxn = -nF(E0 cathode - Eo anode)
dg rxn = -nF(E0 Pb+2/Pb - E0 Mg+2/Mg)
dG rxn = -2 * 96500 * (-0.13 -(-2.36))
dG rxn = -430.39 kJ
b)
Br2(l)+2Cl(aq)2Br(aq)+Cl2(g)
dG rxn = -nFE0 cell
dG rxn = -nF(E0 cathode - Eo anode)
dg rxn = -nF(E0 Br2/Br- - E0 Cl2/Cl-)
dG rxn = -2 * 96500 * (1.09 -(1.36))
dG rxn = 52.110 kJ
c)
MnO2(s)+4H+(aq)+Cu(s)Mn2+(aq)+2H2O(l)+Cu2+(aq)
dG rxn = -nFE0 cell
dG rxn = -nF(E0 cathode - Eo anode)
dg rxn = -nF(E0 MnO2/Mn+2 - E0 Cu+2/Cu)
dG rxn = -2 * 96500 * (1.23 - 0.34)
dG rxn = -171.77 kJ

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.