Your reaction
.. Fe + O2 ---> FexOy
for this reaction..
.. the Fe on the left is in the 0 oxidation state
.. the Fe on the right is in the +(2y/x) oxidation state
.. the O on the left is in the 0 oxidation state
.. the O on the right is in the -2 oxidation state
meaning
.. the O is reduced... . . (it's reduced in oxidation state)
.. the Fe is oxidized.. . .(oxidation state increased)
this is a REDOX reaction
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AND.. it's also a synthesis reaction.. (aka combination reaction)
The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
Answer:
Phase changes that require a loss in energy are condensation and freezing.
Explanation:
it's Lithium or Li
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Lambda = h\ Mv
lambda = 6.624 x 10^-34 / 9.1 x 10^-31 x 2.5 x 10^7
lambda = 2.9 x 10^-11 is your wavelength
