Electrochemistry - Result Question 1
####1. Given,
$$ \begin{aligned} & Co^{3+}+e^{-} \longrightarrow Co^{2+} ; E^{\circ}=+1.81 V \ & Pb^{4+}+2 e^{-} \longrightarrow Pb^{2+} ; E^{\circ}=+1.67 V \ & Ce^{4+}+e^{-} \longrightarrow Ce^{3+} ; E^{\circ}=+1.61 V \ & Bi^{3+}+3 e^{-} \longrightarrow Bi ; E^{\circ}=+0.20 V \end{aligned} $$
Oxidising power of the species will increase in the order
(2019 Main, 12 April I)
(a) $Ce^{4+}<Pb^{4+}<Bi^{3+}<Co^{3+}$
(b) $Bi^{3+}<Ce^{4+}<Pb^{4+}<Co^{3+}$
(c) $Co^{3+}<Ce^{4+}<Bi^{3+}<Pb^{4+}$
(d) $Co^{3+}<Pb^{4+}<Ce^{4+}<Bi^{3+}$
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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 $H^{+} / H _2$ couple
$$ \text { Given, } \begin{aligned} & Co^{3+}+e^{-} \longrightarrow Co^{2+} ; E^{\circ}=+1.81 V \ & Pb^{4+}+2 e^{-} \longrightarrow Pb^{2+} ; E^{\circ}=+1.67 V \ & Ce^{4+}+e^{-} \longrightarrow Ce^{3+} ; E^{\circ}=+1.61 V \ & Bi^{3+}+3 e^{-} \longrightarrow Bi ; E^{\circ}=+0.20 V \end{aligned} $$
Oxidising power of the species increases in the order of $Bi^{3+}<Ce^{4+}<Pb^{4+}<Co^{3+}$.
Higher the emf value, stronger the oxidising power. The maximum value of emf is possessed by $Co^{3+}$. Hence, it has maximum oxidising power. Whereas $Bi^{3+}$ possess the lowest emf value. Hence, it has minimum oxidising power.
