Electrolysis

[atmega-ist]

1. The problem statement, all variables and given/known data

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.


2. Relevant equations



3. The attempt at a solution

(a)

\frac{5.00g Mn}{} \frac{1 mol Mn}{54.94 g Mn} \frac{1 mol rxn}{2 mol Mn} \frac{4 mol e ^{-}}{1 mol rxn} \frac{96485 Coulomb}{1 mol e ^{-}} = 17561. 89 Coulomb

\frac{17561.89 Coulomb}{1 A/ C / s} \frac{1}{6.2A} = 2832.56 s

(b)
3600 s \frac{5.00 Coulomb}{1 s} \frac{1 mol e^{-}}{96485 Coulomb} = .187 mol e-

.187 mol e- \frac{1 mol rxn}{4 mol e^{-}} \frac{2 mol Mn^{2+}}{1 mol rxn} = .0935 mol Mn2+

.21875 mol Mn2+ (initial) - .0935 mol Mn2+ = .716M Mn2+ after electrolysis

for pH:

.0935 mol Mn2+ \frac{4 mol H^{+}}{2 mol Mn^{2+}} \frac{1}{.175 L} = 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.