Ncert Solutions Chemistry Class 12th

$E_{M{n}^{+3}/M{n}^{+2}}^o=1.51V$

the strongest oxidizing agent have the highest reduction potential. So Mn³⁺ is the strongest oxidizing agent.

Cerium exists in two oxidation states (+3) and (+4)

$\begin{array}{l}C{e}^{+4}+e^{-}\to C{e}^{+3}{E}^0=1.61V\\ C{e}^{+3}+3e^{-}\to Ce{E}^0=-2.336V\end{array}$

It exist as Ce+4 and acts like a strong oxidizing agent by gaining electrons

Au + NaCN + O₂ → Na [Au (CN)₂]

$Zn+Na\left[Au{\left(CN\right)}_2\right]\to N{a}_2\left[Zn{\left(CN\right)}_4\right]+Au$

$Ais{\left[Au{\left(CN\right)}_2\right]}^{-}andBis{\left[Zn{\left(CN\right)}_4\right]}^{-2}$

 $A{\mathcal{l}}^{+3}$ → number of electrons = 10

Br → number of electrons = 36

$B{e}^{+2}$ → number of electrons = 2

$C{\mathcal{l}}^{-}$ → number of electrons = 18

$L{i}^{+}$ → number of electrons = 2

$S^{-2}$ → number of electrons = 18

K⁺ → number of electrons = 18

O^-2 → number of electrons = 10

Mg⁺² → number of electrons = 10

O^-2 & Mg⁺² are isoelectronic with Al⁺³

In acidic solution

$3\stackrel{+6}{Mn}{O}_4^{2-}+4H^{+}\to 2\stackrel{+7}{Mn}{O}_4^{-}+\stackrel{+4}{Mn}{O}_2+2H_{2}O$

Difference in oxidation state of Mn is product

= 7 – 4 = 3

Baryte – BaSO₄ [Sulphate based ore]

Galena – PbS

Zinc blende – ZnS

Copper pyrite – CuFeS₂

Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$=\left(0.2*200\right)*10^{-3}-\left(0.1*200\right)*10^{-3}$                

= 20 * 10^-3 moles

Mass of CH₃COOH absorbed = 20 * 10^-3 * 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6}=2g$ of charcoal.