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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:

  1. A Fluorine gas is a strong oxidizing agent.
  2. B Lithium ion is a strong oxidizing agent.
  3. C Copper can be oxidized by nitrate ion.
  4. D Zinc can be oxidized by hydrogen ion.
Verified Solution

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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