Answer:
true
Explanation:
what is the answer? please
Answer: C2H4 + 3O2 → 2CO2 + 2H2O
Explanation:
2Al + 2O2 → 2AlO + O2 Not Balanced Properly: 2Al + O2 = 2AlO
C2H4 + 3O2 → 2CO2 + 2H2O Looks Good
2CH4 + O2 → 2CO + 4H2 Not Correct: CO should be CO2
Ca + O2 → CaOH Not Balanced and No source for the H
Netural atoms lose electrons or gain electrons to form ions. Either positive or negative ions.
Positive ions:
Na ---> Na+ + e-
Negative ions:
Cl + e- ---> Cl-
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
= 
= 1.0 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V