What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C when [Fe2+]= 3.40 M and [Mg2+]= 0.310…
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Question “What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C when [Fe2+]= 3.40 M and [Mg2+]= 0.310…”
What is the cell potential for the reaction
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 69 ∘C when [Fe2+]= 3.40 M and
[Mg2+]= 0.310 M . Express your answer to three significant figures
and include the appropriate units.
Answer
Let’s Find Eo 1st!
Data table:
Eo(Mg2+/Mg (s)) = -2.372V
Eo(Fe2+/Fe(s),) = -0.44V
According to the given reaction/cell notation
Cathode is (Fe2+/Fe(s).
Anode is (Mg2+/Mg(s).
Eocell = Eocathode-Eoanode
= (-0.44) – (-2.372)
= 1.932 V
Balanced reaction has 2 electrons.
Also, n = 2.
use:
E = Eo – (2.303*RT/nF) log [Mg2+]^1/[Fe2+]^1
Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591
The above expression is:
E = Eo – (0.0591/n) log [Mg2+]^1/[Fe2+]^1
E = 1.932 – (0.0591/2) log (0.31^1/3.4^1)
E = 1.932 (-3.075*10-22)
E = 1.963V
Answer: 1.96 V
Conclusion
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