Consider the reaction Mg(s)+F e 2+ (aq)M g 2+ (aq)+Fe(s) at 59 C , where [F e 2+

Question

Consider the reaction Mg(s)+F e 2+ (aq)M g 2+ (aq)+Fe(s) at 59 C , where [F e 2+ ]= 3.80 M and [M g 2+ ]= 0.210 M . Part A What is the value for the reaction quotient, Q , for the cell?

Express your answer numerically. Q =

Part B What is the value for the temperature, T , in kelvins? Express your answer to three significant figures and include the appropriate units. T =

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). n =

Part D Calculate the standard cell potential for Mg(s)+F e 2+ (aq)M g 2+ (aq)+Fe(s) Express your answer to three significant figures and include the appropriate units. E =

Explanation / Answer

The given reaction is

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

A) The reaction quotient is given as

Q = [Mg2+][Fe]/[Mg][Fe2+] = [Mg2+]/[Fe2+] (the concentration of solid species is taken as unity)

Given [Mg2+] = 0.210 M and [Fe2+] = 3.80 M, we have

Q = (0.210 M)/(3.80 M) = 0.0553 0.055 (ans).

B) The temperature of the reaction system is t = 59C. The temperature in the Kelvin scale is T = (59 + 273) K = 332 K (273 K is the absolute zero of temperature) (ans).

C) Write down the half reactions:

Mg (s) -------> Mg2+ (aq) + 2 e- (oxidation)

Fe2+ (aq) + 2 e- ------> Fe (s) (reduction)

Since 2 moles of electrons are involved in each of the processes, hence, the value of n, the number of moles of electrons involved in the redox process is 2 (ans).

D) The standard reduction potential values are obtained from the table:

Mg2+ + 2 e- ------> Mg; E0 = -2.37 V ….(1)

Fe2+ + 2 e- ------> Fe; E0 = -0.44 V ….(2)

The given cell reaction is obtained by subtracting (1) from (2); the standard cell potential is given as

E0cell = E0reduction + E0oxidation = (-0.44 V) + (2.37 V) [E0oxidation is the reverse of E0 in (1)] = 1.93 V (ans).

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