Chemistry · 2021
JEE Main · 25 February 2021, Shift 2 · Q56
Copper reduces NO_3^- into NO and NO_2 depending upon the concentration of HNO_3 in solution. (Assuming fixed [ Cu^2+] and P_NO = P_NO_2 ), the HNO_3…
Copper reduces $\displaystyle \mathrm{NO}_3^{-}$into NO and $\displaystyle \mathrm{NO}_2$ depending upon the concentration of $\displaystyle \mathrm{HNO}_3$ in solution. (Assuming fixed $\displaystyle \left[\mathrm{Cu}^{2+}\right]$ and $\displaystyle \mathrm{P}_{\mathrm{NO}}=\mathrm{P}_{\mathrm{NO}_2}$ ), the $\displaystyle \mathrm{HNO}_3$ concentration at which the thermodynamic tendency for reduction of $\displaystyle \mathrm{NO}_3^{-}$into NO and $\displaystyle \mathrm{NO}_2$ by copper is same is $\displaystyle 10^x \mathrm{M}$. The value of $\displaystyle 2 x$ is \_\_\_\_. (Rounded-off to the nearest integer) [Given, $\displaystyle \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{NO}_3^{-} / \mathrm{NO}}^{\circ}=0.96 \mathrm{~V}, \mathrm{E}_{\mathrm{NO}_3^{-} / \mathrm{NO}_2}^{\circ}=0.79 \mathrm{~V}$ and at $\displaystyle 298$ K , $\displaystyle \left.\frac{\mathrm{RT}}{\mathrm{F}}(2.303)=0.059\right]$
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.