A voltaic cell employs the following redox reaction: 2Fe3+(aq)+3Mg(s)2Fe(s)+3Mg2

Question

A voltaic cell employs the following redox reaction: 2Fe3+(aq)+3Mg(s)2Fe(s)+3Mg2+(aq) Calculate the cell potential at 25 C under each of the following conditions. You may want to reference ( pages 881 - 886) section 18.6 while completing this problem. Part A standard conditions Express your answer in units of volts. Ecell = V Part B [Fe3+]= 1.1×103 M ; [Mg2+]= 1.90 M Express your answer in units of volts. Ecell = V Part C [Fe3+]= 1.90 M ; [Mg2+]= 1.1×103 M Express your answer in units of volts. Ecell = V

Explanation / Answer

2Fe3+(aq)+3Mg(s)2Fe(s)+3Mg2+(aq)

Fe3+ + 3e- Fe                            E(red) = -0.04 V reduction at cathode

Mg Mg2+ + 2e-                         E(red) = -2.36 V oxidation at anode

Balance the equations

2Fe3+ + 6e- 2 Fe                                    E(red) = -0.04 V reduction at cathode

3Mg 3Mg2+ + 6e-                     E(red) = -2.36 V oxidation at anode

PART –A

E0(cell) = E0(cathode) – E0(anode)

E0(cell) = -0.04 V - (-2.36)

E0(cell) = 2.32 V

Part B

[Fe3+]= 1.1×103 M ; [Mg2+]= 1.90 M

Nernst equation, Ecell = E0cell- 0.059 /n log[Mg2+]3/[Fe3+]2

= 2.32 V –{0.059 / 6 log [1.90]3/[1.1 x 10-3]2}

= 2.32 V –{0.059 / 6 log [1.90]3/[1.1 x 10-3]2}

= 2.25 V

PART-C

[Fe3+]= 1.90 M ; [Mg2+]= 1.1×103 M

Ecell = E0cell- 0.059 /6 log[1.1×103]3/[1.90]2

      = 2.41 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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