Are My Electrochemistry Homework Answers Correct?

[mikesown]

This is a multi-question homework which I want to see if I understand fully. I have attempted as much as I can of it, and would appreciate it if members here could tell me if my answers are correct and if not could guide me in the direction. Thanks in advance.

1. Balance the following reaction that occurs in acidic solution
Question: $$S^{-2} + NO_{3}^{-1} -> NO_{2} + S_{8}$$
Answer(correct?): $$32H^{+} + 16NO_{3}^{-} + 8S^{-2} -> S_{8} + 16NO_{2} + 16H_{2}O$$

2.
$$Mg^{+2} + 2e^{-} -> Mg$$ E cell = -2.38V
$$Ag^{+1} + 1e^{-} -> Ag$$ E cell = 0.80V
a. Write the oxidation half reduction, reduction half reduction, and the net reaction and calculate E cell.
Answer(correct?):
Oxidation: $$Mg -> Mg^{+2} + 2e^{-}$$ 2.38V
Reduction: $$Ag^{+1} + 1e^{-} -> Mg$$ .80V
Net: $$Mg+2Ag^{+1} -> Mg^{+2} + 2Ag$$ 3.18V
b. Sketch the above voltiac cell and label the anode, cathode, and salt bridge. Write the formulas of two approciriate electrolytes. Label which electrode gains mass and which one looses mass.

(my best attempt at an ASCII rendition of the drawing)
Salt bridge
|
/------------------\
| |
| | | |
| | | |
| | | |
| | | |
____ ______
#1 #2

Mg in beaker#1(solution), anode, not sure about electrolyte!, Ag+1 in beaker #2, cathode, not sure about electrolyte. Second beaker electrolyte gains mass(because it's being reduced). First beaker looses mass. Not sure about electrolyte for salt bridge.

c) Determine the free energy value for the above Voltaic cell:
-6.14*10^(5) J

d) Find the numerical value of the equilibrium constant
2.71*10^(107) (Is this right? Seems large!)

e) Calculate the cell voltage if the SIlver electrolyte molarity changes to .10M while the magnesium electrolyte molarity is changed to .002M.
Didn't really know. Set up the following:
E cell = E(degree) cell * .0592/2 * log Q

3. A current of 10.0 amperes flows for 2.00 hours through an electrolytic cell containing a molten salt of metal X. The result is the decomposition of .250 moles of metal X at the cathode. The oxidation state of X in the molten state is?

I found the number of mol e-, which was 7.46*10^(-1), but I didn't know where to go from here.

Any help would be _greatly_ appriciated!

Thanks!

 

[chemisttree]

3. A current of 10.0 amperes flows for 2.00 hours through an electrolytic cell containing a molten salt of metal X. The result is the decomposition of .250 moles of metal X at the cathode. The oxidation state of X in the molten state is?

I found the number of mol e-, which was 7.46*10^(-1), but I didn't know where to go from here.

Any help would be _greatly_ appriciated!

Thanks!

.250 moles of metal X and 0.75 moles of e-. How many e-'s per metal ion?

 

[Blueshift5]

1. Your answer for the first question is correct. You have balanced the equation by adding 32 H+ ions and 16 H2O molecules to the left side of the equation.

2.
a. Your answers for the oxidation and reduction half reactions are correct. However, the net reaction should be written as Mg + 2Ag+ -> Mg2+ + 2Ag, with a cell potential of 3.18V. Also, in the reduction half reaction, Ag should have a +1 charge, not a -1 charge.
b. Your drawing is correct. The anode is the electrode where oxidation occurs, so it is where Mg is located. The cathode is where reduction occurs, so it is where Ag is located. The electrolyte in beaker #1 could be Mg(NO3)2 and the electrolyte in beaker #2 could be AgNO3. The electrolyte in the salt bridge could be KNO3 or NaNO3. The electrode that gains mass is the cathode, and the electrode that loses mass is the anode.
c. The free energy value should be -6.14*10^5 J.
d. The numerical value of the equilibrium constant is 2.71*10^107, which is a very large number. This indicates that the reaction strongly favors the products.
e. To calculate the cell voltage, you can use the Nernst equation: Ecell = Ecell° - (0.0592/2)log(Q). Q is the reaction quotient, which is [Ag+]^2/[Mg2+]^2. With the new concentrations, Q will change and therefore the cell voltage will change.

3. To determine the oxidation state of X in the molten state, you can use the Faraday's law of electrolysis: moles of substance = (current * time)/(Faraday's constant * number of electrons transferred). In this case, the number of electrons transferred is 2, so you can solve for the moles of X and then use the mole ratio to determine the oxidation state.