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

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

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

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

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

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