QJEEM21Ch28d1: JEE Main 2021 Chemistry - Class 12 JEEM21Ch28d1 Solution: Copper reduces NO into NO and NO 2 depending upon the…

Cell-I $(HN{O_3} \to NO)$

$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$

${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$

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

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

Cell-II $(HN{O_3} \to N{O_2})$

$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$

${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$

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

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

Now, ${E_1} = {E_2}$

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

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

$= {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$

$= {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$ [ $\because$ ${p_{NO}} = {p_{N{O_2}}}$ ]

$= {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$

Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$

So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$

$= {{0.059} \over 6} \times 8\log [HN{O_3}]$

$\log [HN{O_3}] = 2.16$

$[HN{O_3}] = {10^{2.16}}M = {10^x}M$

$\therefore$ $x = 2.16$

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