Consider the reaction Mg(s)+Fe2+(aq)Mg2+(aq)+Fe(s) at 51 C , where [Fe2+]= 2.90

ID: 978656 • Letter: C

Question

Consider the reaction

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

at 51 C , where [Fe2+]= 2.90 M and [Mg2+]= 0.210 M .

Part A

What is the value for the reaction quotient, Q, for the cell?

Express your answer numerically.

Part B

What is the value for the temperature, T, in kelvins?

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

Part C

What is the value for n?

Express your answer as an integer and include the appropriate units (i.e. enter mol for moles).

Part D

Calculate the standard cell potential for

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

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

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

[Fe2+]= 2.90 M and [Mg2+]= 0.210 M

reaction quoteint , Q = [Mg2+] / [Fe2+]

Q = 0.210 / 2.90

Q = 0.0724

b) Temperature T = 51oC = 51 + 273 = 324 K

T = 324 K

c) Oxidation half reaction(at anode) : Mg(s) -----> Mg2+(aq) + 2e- Eored = -2.37 V

reduction half reaction(at cathode) : Fe2+(aq) + 2e- -------> Fe(s) Eored = -0.44 V

so, number of electrons lost or gained , n = 2

n = 2

d) Eocell = Eored(cathode) - Eored(anode) = -0.44 V - (-2.37 V) Standard cell potential Eocell = 1.93 V

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