Answer: as copper has lower electrode potential value than hydrogen, it could be reduced by hydrogen.
Explanation: hydrogen has zero reduction potential while Cu has +0.34V and Al has -1.66 V .
SO in electrochemical series who has most negative or less reduction potential value tends to be a good reducing agent than the other.
Hope it helps...
Answer:
Refer to your periodic table. Lewis dot structures are based off the number of valence electrons an atom has.
Looking at the compounds, we can see that Gallium has three valence electrons in its outer shell and oxygen has six. Oxygen and Gallium are going to share electrons with one another, making a V shape in their diagram.
One Oxygen would make a double bond with a Gallium, leaving one valence electron to another oxygen. That oxygen takes that Final electron. It now has 7 in its outer shell. The remaining Gallium and Oxygen do the same double bond as the one before, leaving the 7 valence electron oxygen with one more electron.
Add 4H2 in the reactant side, that will give you 4H2O in the product side.
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Answer:
The answer is "Option C".
Explanation:
Please find the complete question and its solution in the attached file.
using Hoffman's elimination reaction.
