Three faradays of electricity is passed through molten al2o3, aqueous solution of cuso4 and molten nacl taken in separate electrolytic cells. The amounts of al, cu and na deposited at the cathodes will be in the molar ratio
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Answer:
0.8 moles Na
Explanation:
The overall reactions look like this
2
Al
2
O
3(l]
→
4
Al
(l]
+
3
O
2(g]
↑
⏐
and
2
Na
+
Cl
−
→
2
Na
(s]
+
Cl
2(g]
↑
⏐
⏐
Now, the half-reactions that are of interest to you are
6
−
2
O
→
3
0
O
2
+
12
e
−
Oxygen is being oxidized to oxygen gas.
+
1
Na
+
+
1
e
−
→
0
Na
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