- #1
mikesown
- 16
- 0
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!
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!
Last edited: