MCQ
Phosphine is prepared by the reaction of
- A$P$ and ${H_2}S{O_4}$
- ✓$P$ and $NaOH$
- C$P$ and ${H_2}S$
- D$P$ and $HN{O_3}$
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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 $........$.