Electrochemistry 2 Question 25
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 $1250 \mathrm{~K}$. However, the $\mathrm{N}{2}$ gas contains $1 \mathrm{~mole} %$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below
$$ 2 \mathrm{Cu}(g)+\mathrm{H}{2} \mathrm{O}(g) \longrightarrow \mathrm{Cu}{2} \mathrm{O}(s)+\mathrm{H}_{2}(g) $$
$p_{\mathrm{H}{2}}$ is the minimum partial pressure of $\mathrm{H}{2}$ (in bar) needed to prevent the oxidation at $1250 \mathrm{~K}$. The value of $\ln \left(p_{\mathrm{H}_{2}}\right)$ is
(Given : total pressure $=1$ bar, $R$ (universal gas constant) $=8 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \quad \ln (10)=2.30 \mathrm{Cu}(s)$ and $\mathrm{Cu}_{2} \mathrm{O}(s)$ are mutually immiscible.)
$$ \begin{aligned} & \text { At } 1250 \mathrm{~K}: 2 \mathrm{Cu}(s)+1 / 2 \mathrm{O}{2}(g) \longrightarrow \mathrm{Cu}{2} \mathrm{O}(s) ; \ & \qquad \Delta G^{\ominus}=-78,000 \mathrm{~J} \mathrm{~mol}^{-1} \ & \mathrm{H}{2}(g)+\frac{1}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \ & \Delta G^{\ominus}=-1,78,000 \mathrm{~J} \mathrm{~mol}^{-1} ; G \text { is the Gibbs energy (2018 Adv.) } \end{aligned} $$
Passage Based Questions
Passage I
The electrochemical cell shown below is a concentration cell. $M \mid M^{2+}$ (saturated solution of a sparingly soluble salt, $\left.M X_{2}\right)|| M^{2+}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid M$. The emf of the cell depends on the difference in concetration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 \mathrm{~K}$ is $0.059 \mathrm{~V}$.
(2012)
Show Answer
Solution:
- Given
(i) $2 \mathrm{Cu}(\mathrm{s})+\frac{1}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{Cu}{2} \mathrm{O}(s) ; \quad \Delta G^{\mathrm{o}}=-78000 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-78 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) $\mathrm{H}{2}(g)+\frac{1}{2} \mathrm{O}{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta G^{\mathrm{o}}=-178000 \mathrm{~J} \mathrm{~mol}^{-1}$
$$ =-178 \mathrm{~kJ} \mathrm{~mol}^{-1} $$
So, net reaction is (By (i)-(ii))
$$ 2 \mathrm{Cu}(s)+\mathrm{H}{2} \mathrm{O}(g) \longrightarrow \mathrm{Cu}{2} \mathrm{O}(s)+\mathrm{H}_{2}(g) $$
$\Delta G=100000 \mathrm{~J} / \mathrm{mol}$ 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,
$$ \Delta G^{\mathrm{o}}+R T \ln \left[\frac{p_{\mathrm{H}{2}}}{p{\mathrm{H}_{2} \mathrm{O}}}\right] \geq 0 $$
After putting the values,
$$ \begin{array}{cc} & 10^{5}+8 \times 1250 \ln \left[\frac{p_{\mathrm{H}{2}}}{p{\mathrm{H}{2} \mathrm{O}}}\right] \geq 0 \ \text { or } & 10^{5}+10^{4} \ln \left[\frac{p{\mathrm{H}{2}}}{p{\mathrm{H}{2} \mathrm{O}}}\right] \geq 0 \ \text { or } & 10^{4}\left(\ln p{\mathrm{H}{2}}-\ln p{\mathrm{H}{2} \mathrm{O}}\right) \geq-10^{5} \ \text { or } & \ln p{\mathrm{H}{2}} \geq-10+\ln p{\mathrm{H}{2} \mathrm{O}} \ \text { or } & \ln p{\mathrm{H}{2}} \geq-10+2.3 \log (0.01)\left(\text { as } p{\mathrm{H}{2} \mathrm{O}}=1 %\right) \ \text { so } & \ln p{\mathrm{H}_{2}} \geq-14.6 \end{array} $$
