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: The answer is 167
Explanation: This is because that was right on edg. so yea heart this tho plsss
Answer:
choice 3
Explanation:
Hydrogen bonding occurs when a H is bonded to a highly electronegative atom like F. Towards the upper right corner, most atoms have high electronegativity and small atomic radii, according to periodic trends.
Please give thanks :)
Answer:
d. 0.208 M NaOH
Explanation:
M[NaOH] = 23+16+1= 40g/mol
2.40L = 2.4dm3
M=m/Mv
M=20.0g/40g/mol×2.4dm3
M=20.0g ÷ 96
M= 0.208 M NaOH