POST-LAB EXERCISES Given the following redox reaction: Mg(s) + Fe2+(aq) ? Mg2+(a

Question

POST-LAB EXERCISES
Given the following redox reaction: Mg(s) + Fe2+(aq) ? Mg2+(aq) + Fe(s)

_____ 1. Which of the following statements is True?
A. Fe can reduce Mg2+(aq) ions C. Mg can oxidize Fe2+(aq) ions
B. Mg can reduce Fe2+(aq) ions D. Fe2+(aq) is the reducing agent
_____ 2. If the reaction quotient (Q) in the Nernst equation is equal to 1.0 x 102 then:

A. Ecell = Eocell B. Ecell > Eocell C. Ecell < Eocell

3. Using Table 1 on page 6, calculate the standard cell potential, Eocell, for the following reaction:
2H+(aq) + 2 I–(aq) ? I2(s) + H2(g). Is this reaction spontaneous? Show your work.

4. Using the half-reaction potentials from Table 1 on page 6, calculate the Eocell for the Ni ? Al cell.
For the Ni ? Al cell in question 4 write the following reactions

5. Cathode Half Reaction:
6. Anode Half Reaction:
7. Redox Reaction:
8. Using your answer from question 4, determine the Ecell for a Ni ? Al cell when the concentration of Al3+(aq) is 4.00 M and the concentration of Ni2+(aq) is 2.00 x 10?5 M.
9. Using your bench average of the experimental value for Ecell from Part III and the equations on page 9, calculate Ksp for AgCl or Cu(OH)2

If the Ag concentration is 0.0500 M in the Ag/AgNO3(aq) half-cell, and the Ag concentration is x M in the gCl(s) half-cell, we can represent the cell reaction as Cathode Compartment Anode Compartment Ag (s) Ag (0.0500 M) Ag (s) Ag (x M) (21) 0.0591 Using the Nernst Equation E cell Edell logg 10Q (14) Where E (cell) 0.00 v, n the number of electron needed to reduce Ag ions Ag+(aq) e Ag(s) [Ag product 0.0591. LAg Equation (14) become equation (22) cell log and Q (22) Ag ]reactant 0.0500 Step 1. the [Ag+] can be obtained from equation (22) 0.0591 [Ag where E is your experimentally measured voltage log cell 0.0500 (23) Step 2 the [Cl 1 can be determined from the dilution equation M1V1 M2V2 M-NaCl NaCl disp enced 13 [CH see step 21 avf of mixture) Step 3 once the [Ag and Cl 1 has been obtained it is a simple matter of plugging these values into the Ksp expression for silver chloride

Explanation / Answer

1)
Here Mg is getting oxidised and Fe2+ is getting reduced
So,
Answer:
B. Mg can reduce Fe2+(aq) ions

2)
E = Eo - 0.059 log Q
if Q > 1 , log Q will be positive and E will be less than Eo
Answer:
C. Ecell < Eocell

rest of the questions are not related to this

I am allowed to answer only 1 question at a time

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.