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atmega-ist
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Homework Statement
The net equation for an electrolytic cell is:
2 Mn2+ + 2H2O --> 2 Mn(S) + O2(g) + 4H+(aq)
(a) How long will it take to deposit 5.00g of manganese, using a current of 6.20A?
(b) A current of 5.00A is passed through the cell for 1.00hr. Starting out with 175mL of 1.25M Mn(NO3)2, what is [Mn2+] after electrolysis? What is the pH of the solution, neglecting the H+ originally present? What is the volume of oxygen given off at 757mm Hg and 25 celsius? Assume 100% efficiency and no change in volume during electrolysis.
Homework Equations
The Attempt at a Solution
(a)
[tex]\frac{5.00g Mn}{}[/tex] [tex]\frac{1 mol Mn}{54.94 g Mn}[/tex] [tex]\frac{1 mol rxn}{2 mol Mn}[/tex] [tex]\frac{4 mol e ^{-}}{1 mol rxn}[/tex] [tex]\frac{96485 Coulomb}{1 mol e ^{-}}[/tex] = 17561. 89 Coulomb
[tex]\frac{17561.89 Coulomb}{1 A/ C / s}[/tex] [tex]\frac{1}{6.2A}[/tex] = 2832.56 s
(b)
3600 s [tex]\frac{5.00 Coulomb}{1 s}[/tex] [tex]\frac{1 mol e^{-}}{96485 Coulomb}[/tex] = .187 mol e-
.187 mol e- [tex]\frac{1 mol rxn}{4 mol e^{-}}[/tex] [tex]\frac{2 mol Mn^{2+}}{1 mol rxn}[/tex] = .0935 mol Mn2+
.21875 mol Mn2+ (initial) - .0935 mol Mn2+ = .716M Mn2+ after electrolysis
for pH:
.0935 mol Mn2+ [tex]\frac{4 mol H^{+}}{2 mol Mn^{2+}}[/tex] [tex]\frac{1}{.175 L}[/tex] = 1.07M
pH = -log10(1.07) = -.029 (see below)
I'm not necessarily concerned about the oxygen part but if someone wouldn't mind suggesting a general path to the solution (not the solution itself - maybe just what I need to find first to get started?)
Otherwise, Just curious to know if the rest is correct. The pH is throwing me off... I keep getting a negative number.
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