Question

What is the cell potential for the reaction

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

at 75 ∘C when [Fe2+]= 3.10 *M* and [Mg2+]= 0.110
*M* .

Express your answer to three significant figures and include the appropriate units.

Answer #1

Lets find Eo 1st

from data table:

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

Eo(Fe2+/Fe(s)) = -0.44 V

As per given reaction/cell notation,

cathode is (Fe2+/Fe(s))

anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode

= (-0.44) - (-2.372)

= 1.932 V

Number of electron being transferred in balanced reaction is
2

So, n = 2

we have below equation to be used:

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

T = 75 oC = 348 K

Here:

2.303*R*T/F

= 2.303*8.314*348/96500

= 0.069

So, above expression becomes:

E = Eo - (0.069/n) log {[Mg2+]^1/[Fe2+]^1}

E = 1.932 - (0.069/2) log (0.11^1/3.1^1)

E = 1.932-(-5.006*10^-2)

E = 1.982 V

Answer: 1.98 V

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Express your answer numerically.
Part B:What is the value for the temperature, T, in
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Part B
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Express your answer to three significant figures and include the
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Part C
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Part...

What is the cell potential for the reaction
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What is the value for the reaction quotient, Q, for the
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Part B
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Part C
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what is the cell potential at 25 ∘C if the concentrations are
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