<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
= ?
![[Ni^{2+}_{diluted}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D)
= 
![[Ni^{2+}_{concentrated}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D)
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V
The answer to you question is 3
Answer:
1) high pH is required
2) other ions are precipitated along with the strontium ions
Explanation:
According to the solubility rules all phosphates are insoluble except those of sodium, potassium, and ammonium. This implies that strontium phosphate is insoluble in water. This explains why strontium ions can be precipitated from drinking water supply using phosphate. The main problem with the precipitation of strontium using phosphate is that it usually requires a high pH as the precipitation occurs under very alkaline conditions.
The main reason why the results may not be accurate is that other ions are precipitated along with the strontium such as calcium ions and magnesium ions. This may lead to inaccurate determination of the amount of strontium ions present.
Tetrahedral is the answer
