<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.
<u>Explanation:</u>
We are given:
![E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V](https://tex.z-dn.net/?f=E%5Eo_%7B%28F_2%2FF%5E-%29%7D%3D2.87V%5C%5CE%5Eo_%7B%28Al%5E%7B3%2B%7D%2FAl%29%7D%3D-1.66V)
The substance having highest positive
potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Aluminium will undergo oxidation reaction and will get oxidized.
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the
of the reaction, we use the equation:
![E^o_{cell}=E^o_{cathode}-E^o_{anode}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7Bcathode%7D-E%5Eo_%7Banode%7D)
![E^o_{cell}=2.87-(-1.66)=4.53V](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3D2.87-%28-1.66%29%3D4.53V)
Hence, the standard electrode potential of the cell is 4.53 V.
a is the answer because all of the other answers are wrtong
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<span>manganese (Mn)
.tellurium (Te)
.chlorine (Cl).
<span>xenon (Xe).</span></span>
Answer:
A
Explanation:
No temperature change was observed, hence the change is neither exothermic nor endothermic. Hence the answer is A.