Copper reduces NO$_3^ -$ into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO = PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO$_3^ -$ into NO and NO2 by copper is same is 10x M. The value of 2x is _______. (Rounded off to the nearest integer)
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
Answer (integer)
1
Solution
<p>Cell-I $(HN{O_3} \to NO)$</p>
<p>$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$</p>
<p>$${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$$</p>
<p>$\because$ $E_1^o = 0.96 - ( - 0.34) = 1.3\,V$</p>
<p>${E_1} = 1.3 - {{0.059} \over 6}\log {Q_1}$</p>
<p>Cell-II $(HN{O_3} \to N{O_2})$</p>
<p>$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$</p>
<p>$${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$$</p>
<p>$\because$ $E_2^o = 0.79 - ( - 0.34)\,V = 1.13\,V$</p>
<p>${E_2} = 1.13 - {{0.059} \over 2}\log {Q_2}$</p>
<p>Now, ${E_1} = {E_2}$</p>
<p>$1.3 - {{0.059} \over 6}\log {Q_1} = 1.13 - {{0.059} \over 2}\log {Q_2}$</p>
<p>$$0.17 = {{0.059} \over 6}[\log {Q_1} - 3\log {Q_2} - = {{0.059} \over 6}\log {{{Q_1}} \over {{Q_2}}}$$</p>
<p>$$ = {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$$</p>
<p>$$ = {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$$ [$\because$ ${p_{NO}} = {p_{N{O_2}}}$]</p>
<p>$= {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$</p>
<p>Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$</p>
<p>So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$</p>
<p>$= {{0.059} \over 6} \times 8\log [HN{O_3}]$</p>
<p>$\log [HN{O_3}] = 2.16$</p>
<p>$[HN{O_3}] = {10^{2.16}}M = {10^x}M$</p>
<p>$\therefore$ $x = 2.16$</p>
<p>$\Rightarrow 2x = 2 \times 2.16 = 4.32 \simeq 4$</p>
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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