Doubt from electrochemistry

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Is it BD?

Sorry I made a mistake in calculation.

AC I got

I'm getting only B....

How can B be an answer
\displaystyle E=E^0-\frac{0.059}{2}log \left(\frac{[H^+]^4×P_{O_2}}{[H_2O]^2} \right)
Here E^0=-1.23

Here [H^+]=10^{-7} P_{O_2}=1bar
See this @Abhishek_2020_5

Sorry, I don't get what are you trying to say, can you explain in layman's terms....

About how much you get E for B option.

idk ans given is B C ...But that could be wrong also the pH must be maintained at 7 by the reactions at cathode and anode

I got now
Send the solution in a minute.

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