.07! you divide 7 by 100%
<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
Answer:
Before you use log you have to multiply the answer of 8.66 mol by 6.02⋅1023 by 3 because the answer of 8.66 mol by 6.02⋅1023 gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present.
Explanation: