Consider the reaction Mg(s) + Fe^2+ (aq) rightarrow Mg^2+ (aq) + Fe(s) at 87 deg

Question

Consider the reaction Mg(s) + Fe^2+ (aq) rightarrow Mg^2+ (aq) + Fe(s) at 87 degree C, where [Fe^2+] = 3.50 M and [Mg^2+] = 0.310 M. What is the value for the reaction quotient, Q, for the cell? Express your answer numerically. What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units. What is the value for n? Express your answer as an integer and include the appropriate units (i.e. enter mol for moles). Calculate the standard cell potential for Mg(s) + Fe^2+ (aq) rightarrow Mg^2+ (aq) + Fe(s) Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Mg2+(aq) + 2 e– Mg(s) E = –2.37 V; oxidation will take place at anode

Fe2+(aq) + 2e– Fe(s) E = –0.44 V ; reduction will take place at cathode

PART D:

E0cell = E0cathode - E0anode = - 0.44 -(-2.37) = 1.93 Volts

PART C:

n = no. of electron transfer from Mg to Fe2+ = 2

n =2

PART B:

T = 870C = 273 + 870 C = 360 K

PART A:

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

Q ={PRODUCT][REACTANT]

WHERE;

[Mg2+] = [ 0.310]

[Fe2+] = [0.350]

Q = [Mg2+] [Fe2+] = [ 0.310] [0.350] = 0.88 is the answer

if the Q = 0.314 as given above, this means there is a printing mistake

for Q = 0.314 value of  Mg2+= 0.110M instead of 0.310 given in question

if this is the case then Q = 0.110.350 = 0.314

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