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Topic: Voltaic cells: mass reduced at the cathode  (Read 8146 times)

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Offline brandnewrocks13

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Voltaic cells: mass reduced at the cathode
« on: October 31, 2008, 07:32:13 PM »
So I tried this problem and am doing something wrong.  Here is the question:


During the discharge of an alkaline battery, 11.0 g of Zn are consumed at the anode of the battery.  What mass of MnO2 is reduced at the cathode during this discharge.

So I came up with this redox rx:

Zn(s) + MnO2(s) + 4H+ --> Zn2+ + Mn2+ +2H20

then I tried to use the stoichiometry to go from grams Zn to grams MnO4
I set it up like this:
(11.0 g Zn)*(1 mol Zn)/(65.39 g Zn)*(1 mol MnO2)/(1mol Zn)*(86.938 g MnO2)/(1 mol MnO2) 


the answer i came out with was 14.6 g, which isn't the right answer on my online hw.  I just am not sure what i'm doing right.  Can someone help me out please!  :)

Offline Borek

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Re: Voltaic cells: mass reduced at the cathode
« Reply #1 on: October 31, 2008, 07:55:57 PM »
Check if you have correct products.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline brandnewrocks13

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Re: Voltaic cells: mass reduced at the cathode
« Reply #2 on: October 31, 2008, 08:01:25 PM »
how do i know if i have the right products? 
I just looked on a standard reduction potentials chart that i've been using for my class and it listed those equations.


Offline macman104

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Re: Voltaic cells: mass reduced at the cathode
« Reply #3 on: October 31, 2008, 08:06:48 PM »
Equation is correct, but your molar masses are incorrect (well at least 1).

Err...sorry, I read magnesium not manganese.

Offline brandnewrocks13

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Re: Voltaic cells: mass reduced at the cathode
« Reply #4 on: October 31, 2008, 08:07:42 PM »
i figured it out.. I justed needed to have a 2:1 ratio for the MnO2 and zinc.  I must've had my equation wrong, but I got it to work out.  thanks for the help :)

Offline macman104

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Re: Voltaic cells: mass reduced at the cathode
« Reply #5 on: October 31, 2008, 08:14:05 PM »
i figured it out.. I justed needed to have a 2:1 ratio for the MnO2 and zinc.  I must've had my equation wrong, but I got it to work out.  thanks for the help :)
Time to take mac back to school, because I can't understand why you would need a 2:1 ratio...

Offline Borek

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