CUET · CHEMISTRY · PYQ PAPER 2025
Based on the standard electrode potential given in the table
| Half-cell reaction | (E./V) at 25.C) |
| \(F _2(g)+2 e ^{-} \rightarrow 2 F^{-}\) | \(2.87\) |
| \(NO _3^{-}+4 H ^{+}+3 e ^{-} \rightarrow NO ( g )+2 H _2 O\) | \(0.97\) |
| \(Cu ^{2+}+2 e ^{-} \rightarrow Cu ( s )\) | \(0.34\) |
| \(2 H ^{+}+2 e ^{-} \rightarrow H _2(g)\) | 0.00 |
| \(Zn ^{2+}+2 e ^{-} \rightarrow Zn ( s )\) | \(-0.76\) |
| \(L i ^{+}+ e ^{-} \rightarrow L i ( s )\) | \(-3.05\) |
Choose the wrong / incorrect statement from the options given below:
- A Fluorine gas is a strong oxidizing agent.
- B Lithium ion is a strong oxidizing agent.
- C Copper can be oxidized by nitrate ion.
- D Zinc can be oxidized by hydrogen ion.
Answer & Solution
Correct Answer
(B) Lithium ion is a strong oxidizing agent.
Step-by-step Solution
Detailed explanation
The standard reduction potential of \(L i ^{+}\) is \(-3.05\) V. A very negative standard reduction potential indicates a very weak oxidizing agent. Therefore, Lithium ion is a very weak oxidizing agent, not a strong one. The incorrect statement is: Lithium ion is a strong…
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