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Given : gathered E _ 1 2 Cl ₂ / Cl ⁻ ^ o =1.36 ~V , E _ Cr ³⁺ / Cr ^ o =-0.74 ~V E _ Cr ₂ O ₇²⁻ / Cr ³⁺ ^ o =1.33 ~V , E _ MnO ₄⁻ / Mn ²⁺ ^ o =1.51 ~V gathered The correct order of reducing power of the species ( Cr , Cr ³⁺, Mn ²⁺ . and . Cl ⁻ ) will be:

Options

  1. AMn ²⁺ < Cl ⁻ < Cr ³⁺ < Cr
  2. BMn ²⁺ < Cl ³⁺ < Cl ⁻ < Cr
  3. CCr ³⁺ < Cl ⁻ < Mn ²⁺ < Cr
  4. DCr ³⁺ < Cl ⁻ < Cr < Mn ²⁺

Correct answer

A. Mn ²⁺ < Cl ⁻ < Cr ³⁺ < Cr

Step-by-step solution

Lower the value of reduction potential higher will be reducing power hence the correct order will be Mn ²⁺ < Cl ⁻ < Cr ³⁺ < Cr

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