MCQ
$23\, g$ of $Na$ will react with methyl alcohol to give
- AOne mole of oxygen
- BOne mole of ${H_2}$
- ✓$\frac{1}{2}$ mole of ${H_2}$
- DNone of these
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$E^o_{Fe^{+2}/Fe} =-0.44\,V$ and $ E^o_{H^+ / O_2 / H_2O} =1.23\, V$
Calculate the $E^o_{cell}$ of the corrosion :- ............. $\mathrm{V}$