हिंदी

Identify the strongest oxidising agent. NaX++eX−⟶Na; E0 = −2.714 V PtX2++2eX−⟶Pt; E0 = +1.200 V IX2+2eX−⟶2IX−; E0 = + 0.535 V CoX2++2eX−⟶Co; 0 = −0.280 V -

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प्रश्न

Identify the strongest oxidising agent.

\[\ce{Na^+ + e^- -> Na}\]; E0 = −2.714 V

\[\ce{Pt^{2+} + 2e^- -> Pt}\]; E0 = +1.200 V

\[\ce{I2 + 2e^- -> 2I^-}\]; E0 = + 0.535 V

\[\ce{Co^{2+} + 2e^- -> Co}\]; 0 = −0.280 V

विकल्प

  • Na

  • Pt2+

  • I2

  • Sn2+

MCQ

उत्तर

Pt2+

Explanation:

The standard reduction potential of the strongest oxidising agent (Pt2+) is the highest (positive).

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