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Anna11 [10]
3 years ago
8

Current of 250. a flows for 24.0 hours at an anode where the reaction occurring is as follows: mn2+(aq) + 2h2o(l) → mno2(s) + 4h

+(aq) + 2e–what mass of mno2 is deposited at this anode?
Chemistry
2 answers:
Tju [1.3M]3 years ago
7 0
From Faraday's 1st law of electrolysis,
Total electricity passed into system = Q = IT = 250 X 24 X 60 X 60
                                                                       = 2.16 X 10^7 C

We know that, 96500 C = 1 F
∴ 2.16 X 10^7 C =  <span>223.8 F
</span>
Now, number of moles of<span> MnO2  deposited = 223.8/2=111.9 
</span>
Finally, 1 mole of MnO2 ≡ 86.94 g
∴ 111.9 mole of MnO2 ≡  111.9 X 86.94 = 9728 g

Thus, <span>mass of MnO2 that will be deposited at anode = 9728 g</span>
Colt1911 [192]3 years ago
7 0

Answer:

The mass of mno2 is deposited at this anode is 9.732x10³ g

Explanation:

please look at the solution in the attached Word file

Download docx
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<h3>What is theoretical yield?</h3>

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