It condenses very long strings of numbers while retaining the general accuracy of the figure.
<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq)
</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s)
Zn⁰(s) → Zn²⁺(aq) +2e⁻
2 electrons are transferred from atom of Zn⁰ to 2 ions of Li⁺.
<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
The affect of plate movement might have on the size of the ocean basin would be negative and over many millenia it will gradually decrease in size