- 1(current)
Q. No.Consider the values of reduction potential:
\(\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+} ; E^{\circ}=+1.81 \mathrm{~V}\)
\(\mathrm{~Pb}^{4+}+2 e^{-} \rightarrow \mathrm{Pb}^{2+} ; E^{\circ}=+1.67 \mathrm{~V}\)
\(\mathrm{Ce}^{4+}+e^{-} \rightarrow C e^{3+} ; E^{\circ}=+1.61 \mathrm{~V}\)
\( \mathrm{Bi}^{3+}+3 e^{-} \rightarrow \mathrm{Bi} ; E^{\circ}=+0.20 \mathrm{~V}\)
The oxidizing power of the species will increase in the order of:
| 1. | \(C o^{3+}<C e^{4+}<B i^{3+}<P b^{4+}\) |
| 2. | \(\mathrm{Co}^{3+}<P b^{4+}<C e^{4+}<B i^{3+}\) |
| 3. | \(C e^{4+}<P b^{4+}<B i^{3+}<C o^{3+}\) |
| 4. | \(B i^{3+}<C e^{4+}<P b^{4+}<C o^{3+}\) |
Subtopic: Electrochemical Series |
77%
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JEE
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How many elements can liberate H2 from dilute acids?
V, Cr, Mn, Fe, Co, Ni, Cu
1. 5
2. 4
3. 2
4. 6
V, Cr, Mn, Fe, Co, Ni, Cu
1. 5
2. 4
3. 2
4. 6
Subtopic: Electrochemical Series |
61%
From NCERT
JEE
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- 1(current)
Q. No.
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