All inforamtion is here. A voltaic cell employs the following redox reaction: 2F

Question

All inforamtion is here.

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.

Part A

standard conditions

Express your answer in units of volts.

Part B

[Fe3+]= 1.9×103 M ; [Mg2+]= 2.75 M

Express your answer in units of volts.

Part C

[Fe3+]= 2.75 M ; [Mg2+]= 1.9×103 M

Express your answer in units of volts.

All inforamtion is here.

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+]= 1.9×103 M ; [Mg2+]= 2.75 M

Express your answer in units of volts.

Ecell =   V  

Part C

[Fe3+]= 2.75 M ; [Mg2+]= 1.9×103 M

Express your answer in units of volts.

Ecell =   V  

Explanation / Answer

Fe+3 + 3 e- --> Fe Eo = - 0.04 volts

Mg+2 + 2 e- --> Mg Eo = -2.37 volts

Eo for 2Fe3+(aq)+3Mg(s)--> 2Fe(s)+3Mg2+(aq) it has 6 moles of electrons transfered

would be +2.37 - 0.04 = +2.33 volts





Calculate the cell potential at A. [Fe3+]= 1.9×103 M ; [Mg2+]= 2.75M

b)Using nernst equation :

E = Eo - (0.0592 /n) (log Q)

E = 2.33 - (0.0592 /6) (log [products] / [reactants])

E = 2.33 - (0.009867) (log [Mg+2]^3 / [Fe+3]^2)

E = 2.33 - (0.009867) (log [2.75]^3 / [1.9×103]^2)

E = 2.33 - (0.009867) [log (20.79 / 3.61 e-6)]

E = 2.33 - (0.009867) (log 5.78 e6)

E = 2.33 - (0.009867) (6.761)

E = 2.33 - 0.0667

E = 2.2633 Volts

c)
Calculate the cell potential at B. [Fe3+]=2.75 M; [Mg2+]=1.9×103 M

Using nernst equation :

E = Eo - (0.0592 /n) (log Q)

E = 2.33 - (0.0592 /6) (log [products] / [reactants])

E = 2.33 - (0.009867) (log [Mg+2]^3 / [Fe+3]^2)

E = 2.33 - (0.009867) (log [1.9×103]^3 / [2.75]^2)

E = 2.33 - (0.009867) [log (6.859 e-9 / 7.5625)]

E = 2.33 - (0.009867) [log (9.069 e-10)]

E = 2.33 - (0.009867) (-9.04)

E = 2.33 - (-0.089)

E = 2.33 + 0.089

Eo = 2.419 volts

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