Discussão:Metal alcalino
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Addition to the article "Metal alcalino"
The sentence "2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑" describes an irreversible process, that is:
- 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑ 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g) ( 2 LiOH (aq) + H2 (g) → 2 Li (s) + 2 H2O (l) )
- N.B.
- The sentence "NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl ↓" is another example of the sentences that describe the irreversible chemical processes.
- NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl ↓ NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (s) ( NaNO3 (aq) + AgCl (s) → NaCl (aq) + AgNO3 (aq) )
The sentence "2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑" describes a redox process, that is:
- 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑ 2 Li0 (s) + 2 H+12O (l) → 2 Li+1OH (aq) + H02 (g)
The sentence "2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g)" may be inferred from the following sentences:
- Li (s) → Li+(aq) + e−
- 2 H+(aq) + 2 e− → H2 (g)
- H2O (l) = H+(aq) + OH−(aq)
- LiOH (aq) = Li+(aq) + OH−(aq)
- Inference
- Li (s) → Li+(aq) + e− 2 H+(aq) + 2 e− → H2 (g) H2O (l) = H+(aq) + OH−(aq) LiOH (aq) = Li+(aq) + OH−(aq)
- 2 Li (s) → 2 Li+(aq) + 2 e− 2 H+(aq) + 2 e− → H2 (g) H2O (l) = H+(aq) + OH−(aq) LiOH (aq) = Li+(aq) + OH−(aq)
- 2 Li (s) + 2 H+(aq) + 2 e− → 2 Li+(aq) + 2 e− + H2 (g) 2 H2O (l) = 2 H+(aq) + 2 OH−(aq) 2 LiOH (aq) = 2 Li+(aq) + 2 OH−(aq)
- 2 Li (s) + 2 H+(aq) → 2 Li+(aq) + H2 (g) 2 H+(aq) = 2 H2O (l) − 2 OH−(aq) 2 Li+(aq) = 2 LiOH (aq) − 2 OH−(aq)
- 2 Li (s) + 2 H2O (l) − 2 OH−(aq) → 2 LiOH (aq) − 2 OH−(aq) + H2 (g)
- 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g)
The sentence "2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑" may be inferred from the following sentences:
- Li (s) → Li+(aq) + e− & Eox = −3.045 V
- 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V
- H2O (l) = H+(aq) + OH−(aq)
- LiOH (aq) = Li+(aq) + OH−(aq)
- Inference
- Li (s) → Li+(aq) + e−& Eox = −3.045 V 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V H2O (l) = H+(aq) + OH−(aq) LiOH (aq) = Li+(aq) +
- + OH−(aq)
- 2 Li (s) + 2 H+(aq) + 2 e− → 2 Li+(aq) + 2 e− + H2 (g) & Eredox = Ered − Eox = (0 − −3.045) V = +3.045 V 2 H2O (l) = 2 H+(aq) + 2 OH−(aq)
- 2 LiOH (aq) = 2 Li+(aq) + 2 OH−(aq)
- 2 Li (s) + 2 H+(aq) + 2 e− → 2 Li+(aq) + 2 e− + H2 (g) & Eredox = Ered − Eox = (0 − −3.045) V = +3.045 V 2 H2O (l) = 2 H+(aq) + 2 OH−(aq)
- 2 Li (s) + 2 H+(aq) → 2 Li+(aq) + H2 (g) & Eredox = 3.045 V 2 H+(aq) = 2 H2O (l) − 2 OH−(aq) 2 Li+(aq) = 2 LiOH (aq) − 2 OH−(aq)
- 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g) & Eredox > 0 V
- 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 ↑
The sentence "2 Li (s) + 2 H2O (l) → T = 298.15 K 2 LiOH (aq) + H2 (g) & Eredox = 3.045 V" contains the adverbial clause "T = 298.15 K" and may be inferred from the following sentences:
- T = 298.15 K Li (s) → Li+(aq) + e− & Eox = −3.045 V
- T = 298.15 K 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V
- T = 298.15 K H2O (l) = H+(aq) + OH−
- T = 298.15 K LiOH (aq) = Li+(aq) + OH−(aq)
- Inference
- ( T = 298.15 K Li (s) → Li+(aq) + e− & Eox = −3.045 V ) ( T = 298.15 K 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V )
- ( T = 298.15 K H2O (l) = H+(aq) + OH−(aq) ) ( T = 298.15 K LiOH (aq) = Li+(aq) + OH−(aq) )
- T = 298.15 K ( Li (s) → Li+(aq) + e− & Eox = −3.045 V 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V H2O (l) = H+(aq) + OH−(aq)
- LiOH (aq) = Li+(aq) + OH−(aq) )
- T = 298.15 K ( Li (s) → Li+(aq) + e− & Eox = −3.045 V 2 H+(aq) + 2 e− → H2 (g) & Ered = 0 V H2O (l) = H+(aq) + OH−(aq)
- T = 298.15 K 2 Li (s) + 2 H2O (l) → 2 LiOH (aq) + H2 (g) & Eredox = 3.045 V
- 2 Li (s) + 2 H2O (l) → T = 298.15 K 2 LiOH (aq) + H2 (g) & Eredox = 3.045 V