Answer:
pH = 2.059
Explanation:
At the Cathode:
The reduction reaction is:

At the anode:
At oxidation reaction is:

The overall equation for the reaction is:

The overall cell potential is:



Using the formula for the Nernst equation:

where;
E = 0.66
(Zn^2+)=0.22 M
Then
![0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )](https://tex.z-dn.net/?f=0.66%20%3D0.76-%20%28%20%5Cdfrac%7B0.0591%7D%7B2%7D%29log%20%5Cbigg%20%28%20%5Cdfrac%7B%5BZn%5E%7B2%2B%7D%20%5D%20PH_2%7D%7B%5BH%5E%2B%5D%5E2%7D%20%5Cbigg%20%29)
![0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )](https://tex.z-dn.net/?f=0.66%20%3D0.76-%200.02955%20%2A%20log%20%5Cbigg%20%28%20%5Cdfrac%7B0.22%2A0.87%7D%7B%5BH%5E%2B%5D%5E2%7D%20%5Cbigg%20%29)
3.4 = log ( 0.1914) - 2 log [H⁺]
3.4 = -0.7180 - 2 log [H⁺]
3.4 + 0.7180 = - 2 log [H⁺]
4.118 = - 2 log [H⁺]
pH = log [H⁺] = 4.118/2
pH = 2.059
12 x 10^23 atoms, i hope this helps
Explanation:
the answer and explanation is in the picture
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