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

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.

Part A

standard conditions

Part B

[Fe3+]= 1.7×103 M ; [Mg2+]= 1.95 M

Part C

[Fe3+]= 1.95 M ; [Mg2+]= 1.7×103 M

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

Ecell =   V

Part B

[Fe3+]= 1.7×103 M ; [Mg2+]= 1.95 M

Ecell =   V  

Part C

[Fe3+]= 1.95 M ; [Mg2+]= 1.7×103 M

Explanation / Answer

A) Eo cell = Eo ( reductin half) - Eo ( oxidation half)

     = Eo ( Fe3+/Fe) - Eo ( Mg2+/Mg)

        = -0.036 - (-2.37)

        = 2.334 V

b) Q cell= [Mg2+]^3 / [Fe3+]^2

      = ( 1.95^3) / ( 1.7 x 10^ -3)^2 = 2.566 x 10^ 6

E cll = Eo cell - ( RT/nF) ln Q         where n= number of electrons involved per reaction = 6

Ecell = 2.334 - ( 8.314 x 298 / 6 x 96485) ln ( 2.566 x 10^ 6)

     = 2.27

c) Q = ( 0.0017^3) / ( 1.95^2) = 1.29 x 10^ -9

E cell = 2.334 - ( 8.314 x 298 / 6 x 96485 ) ln ( 1.29 x 10^-9)

= 2.42 V

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