M g ( s ) + F e 2 + ( a q ) M g 2 + ( a q ) + F e ( s ) Q is 5.53x10-2 T=318 N=2

Question

Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s)

Q is 5.53x10-2

T=318

N=2

E=1.92V

where E is the potential in volts, E is the standard potential in volts, R 8.314 J/(K mol) is the gas constant T is the temperature in kelvins, s the number of moles of electrons transferred, F-96,500 C/(mol e is the Faraday constant, and Q is the reaction quotient. Substituting each constant into the equation the result is 0.0592 V og10Q Correct Part E What is the cell potential for the reaction Mg(s) +Feet (aq)- Mg (aq) +Fe(s) at 45 C when Fe2+ 3.80 Mand M 0.210 M Express your answer to three significant figures and include the appropriate units. alue Units Submit Hints ML Answers Give Up Review Part Incorrect Try Again; 2 attempts remaining Eeedback Continue D 8:22 AM 4/22/2017

Explanation / Answer

E)

Lets find Eo 1st

from data table:

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

Eo(Fe2+/Fe(s)) = -0.440

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

here:

cathode is (Fe2+/Fe(s))

anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode

= (-0.440) - (-2.372)

= 1.932 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

Use:

E = Eo - (2.303*RT/nF) log {[Mg2+]^1/[Fe2+]^1}

2.303*R*T/n = 2.303*8.314*318.0/F= 0.0631

E = Eo - (0.0631/n) log {[Mg2+]^1/[Fe2+]^1}

E = 1.932 - (0.0631/2) log (0.210^1/3.80^1)

E = 1.932-(-0.040)

E = 1.972 V

Answer: 1.97 V

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