Electrochemistry 1 Question 1
1. Given,
$ \begin{aligned} & \mathrm{Co}^{3+}+e^{-} \longrightarrow \mathrm{Co}^{2+} ; E^{\circ}=+1.81 \mathrm{V} \\ & \mathrm{Pb}^{4+}+2 e^{-} \longrightarrow \mathrm{Pb}^{2+} ; E^{\circ}=+1.67 \mathrm{V} \\ & \mathrm{Ce}^{4+}+e^{-} \longrightarrow \mathrm{Ce}^{3+} ; E^{\circ}=+1.61 \mathrm{V} \\ & \mathrm{Bi}^{3+}+3 e^{-} \longrightarrow \mathrm{Bi} ; E^{\circ}=+0.20 \mathrm{V} \end{aligned} $
Oxidising power of the species will increase in the order
(2019 Main, 12 April I)
(a) $\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}$
(b) $\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$
(c) $\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}$
(d) $\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}$
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Answer:
Correct Answer: 1. (b)
Solution:
- Key Idea Negative $E^{\circ}$ means that redox couple is weaker oxidising agent than $H^+ / H_2$ couple. Positive $E^{\circ}$ means that redox couple is a stronger oxidising agent than $\mathrm{H}^{+} / \mathrm{H} _{2}$ couple
Given,
$ \begin{aligned} & \mathrm{Co}^{3+}+e^{-} \longrightarrow \mathrm{Co}^{2+} ; E^{\circ}=+1.81 \mathrm{V} \\ & \mathrm{Pb}^{4+}+2 e^{-} \longrightarrow \mathrm{Pb}^{2+} ; E^{\circ}=+1.67 \mathrm{V} \\ & \mathrm{Ce}^{4+}+e^{-} \longrightarrow \mathrm{Ce}^{3+} ; E^{\circ}=+1.61 \mathrm{V} \\ & \mathrm{Bi}^{3+}+3 e^{-} \longrightarrow \mathrm{Bi} ; E^{\circ}=+0.20 \mathrm{V} \end{aligned} $
Oxidising power of the species increases in the order of $\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$.
Higher the emf value, stronger the oxidising power. The maximum value of emf is possessed by $\mathrm{Co}^{3+}$. Hence, it has maximum oxidising power. Whereas $\mathrm{Bi}^{3+}$ possess the lowest emf value. Hence, it has minimum oxidising power.
