JEE Main 2021 — Electrochemistry Question with Solution

Cell-I $$(HN{O}_3\to NO)$$

$$3Cu+2N{O}_3^{-}+8H^{+}\to 3C{u}^{2+}+2NO+4H_{2}O$$

$$Q_1=\frac{{[C{u}^{2+}]}^3\times {(p_{NO})}^2}{{[N{O}_3^{-}]}^2\times {[H^{+}]}^8}$$

$$\because$$ $$E_1^o=0.96-(-0.34)=1.3V$$

$$E_1=1.3-\frac{0.059}{6}\log Q_1$$

Cell-II $$(HN{O}_3\to N{O}_2)$$

$$Cu+2N{O}_3^{-}+4H^{+}\to C{u}^{2+}+2N{O}_2+2H_{2}O$$

$$Q_2=\frac{[C{u}^2]\times {(p_{N{O}_2})}^2}{{[N{O}_3^{-}]}^2\times {[H^{+}]}^4}$$

$$\because$$ $$E_2^o=0.79-(-0.34)V=1.13V$$

$$E_2=1.13-\frac{0.059}{2}\log Q_2$$

Now, $$E_1=E_2$$

$$1.3-\frac{0.059}{6}\log Q_1=1.13-\frac{0.059}{2}\log Q_2$$

$$0.17=\frac{0.059}{6}[\log Q_1-3\log Q_2-=\frac{0.059}{6}\log \frac{Q_1}{Q_2}$$

$$=\frac{0.059}{6}\log \frac{{[C{u}^{2+}]}^3\times {(p_{NO})}^2}{{[N{O}_3^{-}]}^2\times {[H^{+}]}^8}\times \frac{{[N{O}_3^{-}]}^6\times {[H^{+}]}^{12}}{{[C{u}^{2+}]}^3\times {(p_{N{O}_2})}^6}$$

$$=\frac{0.059}{6}\log \frac{{[H^{+}]}^4\times {[N{O}_3^{-}]}^4}{{(p_{N{O}_2})}^4}$$ [$$\because$$ $$p_{NO}=p_{N{O}_2}$$]

$$=\frac{0.059}{6}\log \frac{{[HN{O}_3]}^4}{{(p_{N{O}_2})}^4}$$

Now, $$p_{N{O}_2}\equiv [HN{O}_3]$$

So, $$0.17=\frac{0.059}{6}\log [HN{O}_3{]}^8$$

$$=\frac{0.059}{6}\times 8\log [HN{O}_3]$$

$$\log [HN{O}_3]=2.16$$

$$[HN{O}_3]=10^{2.16}M=10^{x}M$$

$$\therefore$$ $$x=2.16$$

$$\Rightarrow 2x=2\times 2.16=4.32\simeq 4$$