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genyjoannerubiera
16.10.2020 •
Chemistry
Write the half reactions and the balanced equation for the galvanic cell Cu(s)| Cu2+(aq)|| Cu+(aq)| Cu(s). What is the smallest possible integer coefficient of Cu+(aq) in the combined balanced equation?
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Ответ:
Answer and Explanation:
The galvanic cell is:
Cu(s)| Cu²⁺(aq)|| Cu⁺(aq)| Cu(s)
The first two species before the double bar (||) constitute the anodic half reaction (oxidation):
Cu(s) ⇒ Cu²⁺(aq) + 2 e-
The two species after the || constitute the cathodic half reaction (reduction):
Cu⁺(aq) + e- ⇒ Cu(s)
If we multiply the reduction half reaction by 2 (to obtain the same number of electrons than oxidation reaction) and then we add the two half reactions, we obtan the balanced equation:
Reduction (cathode) : 2Cu⁺(aq) +2 e- ⇒ 2Cu(s)
Oxidation (anode) : Cu(s) ⇒ Cu²⁺(aq) + 2 e-
Total equation: 2Cu⁺(aq) + Cu(s) ⇒ 2Cu(s) + Cu²⁺(aq)
Cu(s) is in both reactants side and products side, so we cancel that in both opposite sides to obtain:
2Cu⁺(aq) ⇒ Cu(s) + Cu²⁺(aq)
If we divide the balanced equation into 2, the smallest possible integer coefficient for Cu⁺(aq) is 1:
Cu⁺(aq) ⇒ 1/2 Cu(s) + 1/2 Cu²⁺(aq)
Ответ:
324 g of H₂O - Option B.
Explanation:
TiCl₄ + 2H₂O → TiO₂ + 4HCI
1 mol of TiO₂ needed 2 mol of water to be produced.
9 moles of TiO₂ would need 18 mol of water, to be produced (2x9)
1 mol of water weighs 18 g
18 mol of water weigh (18 . 18)/1 = 324 g