Electrochemistry Result Question 24
24. The surface of copper gets tarnished by the formation of copper oxide. $\mathrm{N}_2$ gas was passed to prevent the oxide formation during heating of copper at 1250 K . However, the $\mathrm{N}_2$ gas contains 1 mole $%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below
$ 2 \mathrm{Cu}(g)+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Cu}_2 \mathrm{O}(s)+\mathrm{H}_2(g) $
$p_{\mathrm{H}_2}$ is the minimum partial pressure of $\mathrm{H}2$ (in bar) needed to prevent the oxidation at 1250 K . The value of $\ln \left(p{\mathrm{H}_2}\right)$ is …….
(Given : total pressure $=1$ bar, $R$ (universal gas constant) $=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \ln (10)=2.30 ,\mathrm{Cu}(s)$ and $\mathrm{Cu}_2 \mathrm{O}(s)$ are mutually immiscible.)
At $1250 \mathrm{~K}: 2 \mathrm{Cu}(s)+1 / 2 \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{Cu}_2 \mathrm{O}(s)$;
$ \Delta G^{\ominus}=-78,000 \mathrm{~J} \mathrm{~mol}^{-1} $
$\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ;$ $\Delta G^{\ominus}=-1,78,000 \mathrm{~J} \mathrm{~mol}^{-1} ; $
$G$ is the Gibbs energy
(2018 Adv.)
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Answer:
Correct Answer: 24. $( -14.16 )$
Solution:
- Given
(i)
$ \begin{aligned} 2 \mathrm{Cu}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{Cu}_2 \mathrm{O}(\mathrm{s}) ; \quad \Delta G^{\mathrm{o}} & =-78000 \mathrm{~J} \mathrm{~mol}^{-1} \\ & =-78 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
(ii)
$ \begin{aligned} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) ; \Delta G^{\mathrm{o}} & =-178000 \mathrm{~J} \mathrm{~mol}^{-1} \\ & =-178 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
So, net reaction is (By (i)-(ii))
$ 2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Cu}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2(\mathrm{~g}) \text {; } $
$\Delta G=100000 \mathrm{~J} / \mathrm{mol}$ or $10^5 \mathrm{~J} / \mathrm{mol}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} $
Now, for the above reaction
$ \Delta G=\Delta G^{\mathrm{o}}+R T \ln \left[\frac{p_{\mathrm{H}2}}{p{\mathrm{H}_2 \mathrm{O}}}\right] $
and to prevent above reaction,
$ \Delta G \geq 0 $
So, $\quad \Delta G^{\mathrm{o}}+R T \ln \left[\frac{p_{\mathrm{H}2}}{p{\mathrm{H}_2 \mathrm{O}}}\right] \geq 0 $
After putting the values,
$ 10^5+8 \times 1250 \ln \left[\frac{p_{\mathrm{H}2}}{p{\mathrm{H}_2 \mathrm{O}}}\right] \geq 0 $
or, $\quad 10^5+10^4 \ln \left[\frac{p_{\mathrm{H}2}}{p{\mathrm{H}_2 \mathrm{O}}}\right] \geq 0 $
or, $\quad 10^4\left(\ln p_{\mathrm{H}2}-\ln p{\mathrm{H}_2 \mathrm{O}}\right) \geq-10^5 $
or, $\quad \begin{gathered} \ln p_{\mathrm{H}2} \geq-10+\ln p{\mathrm{H}2 \mathrm{O}} \\ \ln p{\mathrm{H}2} \geq-10+2.3 \log (0.01)\left(\operatorname{as} p{\mathrm{H}_2 \mathrm{O}}=1 %\right) \\ \geq-10-4.6 \end{gathered} $
so, $\quad \ln p_{\mathrm{H}_2} \geq-14.6 $
BETA
