Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
=-0.74V[/tex]
=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials, when concentration is 1M.
![E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Thus the potential of the following electrochemical cell is 1.08 V.
Answer:
Standard form: (x+3)^2=1/2(y+3)
f(1) = 29
f(-1) = 5
Explanation:
The standard form of a parabola with a directrix that is horizontal is
(x-h)=4(P)(y-k)
Using the vertex form, find the vertex, foci, and the distance from the vertex to the focus or directrix.
It's easier to use the vertex form to plug in values for x.
f(1) = 2((1)+3)^2-3
f(1) = 29
f(-1) = 2((-1)+3)^2-3
f(-1) = 5
214, 84 Po ----Beta decay
Answer:
one-half
Explanation:
cuz for a first order reaction is a half life independent of concentration and constant over time
