MCQ
With cold and dilute sodium hydroxide fluorine reacts to give
- ✓$NaF$ and $O{F_2}$
- B$NaF + {O_3}$
- C${O_2}\;{\rm{and}}\;{O_3}$
- D$NaF + {O_2}$
$2NaOH + 2{F_2} \to 2NaF + O{F_2} + {H_2}O$
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Given : $K _{ sp } Cu ( OH )_2=1 \times 10^{-20}$
$\operatorname{Take} \frac{2.303 RT }{ F }=0.059 \,V$
The reduction potential at $pH =14$ for the above couple is $(-) x \times 10^{-2}\,V$. The value of $x$ is $........$.
