Question

Finally, the I compartment contains

I. $C{O}^{2+}$

II. $N{O}_3^{\ominus }$

III. $K^{\oplus }$

Iv. $A{g}^{\oplus }$

• A. $I$
• B. $I,II$
• C. $I,II,III$
• D. $II,III,IV$

Answer 1

Answer: (B)

$I,II$

	I compartment	II compartment	III compartment
Initial moles	$C{o}^{2+}=0.1M\times 5L=0.5$ $N{O}_3^{⊝}=0.2M\times 5L=1.0$	$K^{\oplus }=0.1M\times 5L=0.5$ $N{O}_3^{⊝}=0.1M\times 5L=0.5$	$A{g}^{\oplus }=0.1M\times 5L=0.5$ $N{O}_3^{⊝}0.1M\times 5L=0.5$
Effect of electrode	Formation of $C{o}^{2+}$ $Co⟶C{o}^{2+}+2e^{-}$ $2e^{-}=2F=1molC{o}^{2+}$ $0.1F=\frac{0.1}{2}=+0.05molC{o}^{+2}$	-	Reduction of $A{g}^{\oplus }$ $A{g}^{\oplus }+e^{-}⟶Ag$ $1F=1e^{-}=1molA{g}^{\oplus }$ $0.1F=-0.1molA{g}^{\oplus }$
Positive ion Movement	Migration of $C{o}^{2+}$ For $0.05Eq.=-\frac{0.05}{2}$ $=-0.025molC{o}^{2+}$ (migrated to II compartment)	$+0.025molC{o}^{2+}$ $=+0.025\times 2Eq.C{o}^{2+}$ $=+0.05Eq.C{o}^{2+}$	$+0.05mol{K}^{\oplus }$ (From II compartment)
		$K^{\oplus }=-0.05Eq.$ or $mol$ (migrated to III compartment)
Negative ion movement	$+0.05mol.N{O}_3^{⊝}$ from III compartment	_	$-0.05molN{O}_3^{⊝}$
Final concetration (M)	$C{o}^{2+}=+0.05-0.025$ $=0.025mol.left$ Total $C{o}^{2+}=$ Initial mol + left mol $=0.5+0.025=0.525$ $\left[C{o}^{2+}\right]=\frac{0.525}{5L}=0.105M$	$\left[C{o}^{2+}\right]=\frac{0.025}{5L}=0.005M$	$\left[A{g}^{\oplus }\right]=\left(\frac{0.5-0.1}{5L}\right)$ $=0.08M$
		$\left[K^{\oplus }\right]=\left(\frac{0.5-0.05}{5L}\right)=0.09M$	$\left[N{O}_3^{⊝}\right]=\left(\frac{0.5-0.05}{5L}\right)$ $=0.09M$
	$\left[N{O}_3^{⊝}\right]=\frac{1+0.05}{5L}=0.210M$	$\left[N{O}_3^{⊝}\right]=0.1M$ (No change)	$\left[K^{\oplus }\right]=\frac{0.05}{5L}=0.01M$

There is movement of $N{O}_3^{\ominus }$ ions from III to II and II to I compartment to maintain electrical neutrality in I compartment since there is increase in positive charge due to the oxidation of Co to $C{o}^{2+}$ in I compartment. Hence, ions present in I compartment are :

$C{O}^{2+}$ and $N{O}_3^{\ominus }$.