Electrochemistry

Electrochemistry

Question

Given, 

straight E subscript Cr to the power of 3 plus end exponent divided by Cr end subscript superscript straight o space equals space minus space 0.74 space straight V semicolon
straight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript straight o space equals space 1.51 space straight V
straight E subscript Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript divided by Cr to the power of 3 plus end exponent end subscript superscript straight o space equals space 1.33 space straight V semicolon
straight E subscript Cl divided by Cl to the power of negative 1 end exponent end subscript superscript straight o space equals space 1.36 space straight V
Based on the data given above, strongest oxidising agent will be

  • Cl

  • Cr3+

  • Mn2+

  • MnO4-

Answer

D.

MnO4-

Higher the SRP, better is an oxidising agent, among the givenstraight E subscript MnO subscript 4 superscript minus divided by Mn to the power of 2 plus end exponent end subscript superscript straight o is highest, hence, MnO4- is a strongest oxidising agent.

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