Question

Consider the reaction
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)
at 65 ∘C , where [Fe2+]= 3.40 M and [Mg2+]= 0.110 M .
Part A
What is the value for the reaction quotient, Q, for the cell?

Part B

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

Part C

What is the value for n?

Part D

Calculate the standard cell potential for

Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)

Answer 1

1.The reaction quotient (Q) for the cell is determined by,

Q = [Mg²⁺]/ [Fe²⁺]

Q = 0.110 M / 3.40 M

Q = 0.03235

2. The value for the temperature (T) is given by,

T = 65 ∘C

= (65 + 273) K

= 338 K

3. The value for n is 2 mole.

4. The standard cell potential is determined by,

Mg²⁺ (aq) + 2 e^- ---> Mg (s) E° = -2.36 V
Fe²⁺ (aq) + 2 e^- ---> Fe (s) E° = -0.44 V

Mg (s) --->Mg²⁺ (aq) + 2 e^- E° = 2.36 V ---------------------------------(1)
Fe²⁺ (aq) + 2 e^- --->Fe (s) E° = -0.44 V --------------------------------(2)

Adding (1) and (2), we get;

Mg (s) + Fe2+ (aq) ---> Mg2+ (aq) + Fe (s) E° = 1.92 V

The standard cell potential is 1.92V