Exercise 19.74 Enhanced-with Feedback You may want to reference (Da pages910-914

Question

Exercise 19.74 Enhanced-with Feedback You may want to reference (Da pages910-914 section 196while completing this problem. A voltaic cell employs the following redox reaction 2 Fe (aq) +3 Mig (s) 2 Fe(s) +3 Mg (aq) Calculate the cell potential at 25 Cunder each of the following conditions. Part A standard conditions Express your answer in units of volts. My Answers Give Up Submit Part B Fer 1.7x10 M Mg 325 M Express your answer in units of volts. Eceu. My Answers Give Up Submit Part C 325 M Mg 1.7x10 3M Fe Express your answer in units of volts.

Explanation / Answer

A)

from data table:

Eo(Mg2+/Mg(s)) = -2.372

Eo(Fe3+/Fe(s)) = -0.040

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Fe3+/Fe(s))

anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode

= (-0.040) - (-2.372)

= 2.332 V

Answer: 2.332 V

B)

Number of electron being transferred in balanced reaction is 6

So, n = 6

Use:

E = Eo - (2.303*RT/nF) log {[Mg2+]^3/[Fe3+]^2}

2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591

E = Eo - (0.0591/n) log {[Mg2+]^3/[Fe3+]^2}

E = 2.332 - (0.0591/6) log (3.25^3/0.0017^2)

E = 2.332-(0.070)

E = 2.262 V

Answer: 2.262 V

C)

Number of electron being transferred in balanced reaction is 6

So, n = 6

Use:

E = Eo - (2.303*RT/nF) log {[Mg2+]^3/[Fe3+]^2}

2.303*R*T/n = 2.303*8.314*298.0/F= 0.0591

E = Eo - (0.0591/n) log {[Mg2+]^3/[Fe3+]^2}

E = 2.332 - (0.0591/6) log (0.0017^3/3.25^2)

E = 2.332-(-0.092)

E = 2.424 V

Answer: 2.424 V

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