same amount of electricity was passed through two cells containing molten al2o3 and molten nacl. if 1.8 of al were liberated in one cell the amount of na liberated in the other cell is
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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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