Exercise 20.70 with feedback Part A standard conditions Express your answer in u

Question

Exercise 20.70 with feedback

Part A

standard conditions

Express your answer in units of volts.

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Correct

Ecell = Ecathode Eanode

In this solution, Mg is more negative; therefore, it was assigned to the cathode. Fe is more positive, and therefore it was assigned to the anode.

Ecell = 0.036 V (2.37 V) = 2.33

Under standard conditions, Ecell is positive. Therefore, this reaction is spontaneous in the forward direction.

Part B

[Fe3+]= 1.5×103 M ; [Mg2+]= 3.20 M

Express your answer in units of volts.

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Part C

[Fe3+]= 3.20 M ; [Mg2+]= 1.5×103 M

Express your answer in units of volts.

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Exercise 20.70 with feedback

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.

You may want to reference( pages 933 - 937) section 20.6 while completing this problem.

Part A

standard conditions

Express your answer in units of volts.

Ecell = 2.33   V  

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Correct

Under standard conditions, the standard cell potential can be calculated using the following formula because the concentration of each ion is assumed to be 1 M:

Ecell = Ecathode Eanode

In this solution, Mg is more negative; therefore, it was assigned to the cathode. Fe is more positive, and therefore it was assigned to the anode.

Ecell = 0.036 V (2.37 V) = 2.33

Under standard conditions, Ecell is positive. Therefore, this reaction is spontaneous in the forward direction.

Part B

[Fe3+]= 1.5×103 M ; [Mg2+]= 3.20 M

Express your answer in units of volts.

Ecell =   V  

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Part C

[Fe3+]= 3.20 M ; [Mg2+]= 1.5×103 M

Express your answer in units of volts.

Ecell =   V  

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Explanation / Answer

2 Fe3+(aq) + 3 Mg(s) 2 Fe(s) + 3 Mg2+(aq)

E0 cell = E0 cathode - E0 anode

E0 cell = -0.036 -(-2.37) = 2.33

E cell = E0 cell - 2.303 RT/nF*log([Fe+2]/[Mg+2])

R = universal constant = 8.314 J

T = absolute temperature in K

n = number of electrons involved

F = faraday = 96500 colombs

Part B

E cell = E0 cell - 2.303 RT/nF*log([Fe+2]/[Mg+2])

Ecell = 2.33 - 2.303*8.314*(25+273)/(6*96500) * log(1.5*10^-3/3.2)

Ecell = 2.363 V

Part C

E cell = E0 cell - 2.303 RT/nF*log([Fe+2]/[Mg+2])

Ecell = 2.33 - 2.303*8.314*(25+273)/(6*96500) * log(3.2/1.5*10^-3)

Ecell = 2.2972 V

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