‘Which reaction would most likely require the use of an inert electrode?
1 answer:
An inert electrode is made up of non reactive metals like Platinum
An inert is electrode is required in one of the half cells or both the cells if there is no solid conducting metal in either half reaction. Or we can say that if in any half reaction neither product nor reactant is in solid phase.
From the given equations it is clear that
a) in first cell we have Cu and Ag in solid phase in either side
b) In second reaction the iron is in aqueous phase on both the side of reactant and product
so for reduction of iron or in cathodic half cell we need inert electrode [which does not participate in reaction]
Hence answer is
c) in third reaction Zn and Sn are in solid phase in either side
d) in fourth reaction Al and Ni are in solid phase in either side
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Answer:
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.
1-Propene has eight sigma bonds and one pi bond.
Explanation:
The lewis structure of 1-Propene is shown below. All sigma bonds are highlighted by blue color and the pi bond is highlighted by red color.
Carbon having four unpaired electrons in its excited state can form four covalent (sigma bonds) bonds. In given structure the orbitals containing unpaired electron in sp³ hybridized C undergo head to head overlap with three hydrogen atoms having single unpaired electron and orbital of other sp² hybridized carbon in same fashion. So, four sigma bonds are formed by sp³ hybridized carbon atom.
The doubly bonded carbon atoms also form sigma bond but two unpaired electrons of these two carbons undergo the formation of pi bond. This pi bond forms when the orbitals overlap along the plane perpendicular to the existing sigma bond resulting in the formation of single pi bond.