The $straight E subscript straight M to the power of 3 plus end exponent divided by straight M to the power of 2 plus end exponent end subscript superscript straight o$ values for Cr, Mn, Fe and Co are – 0.41, +1.57, + 0.77 and +1.97 V respectively. For which one of these metals the change in oxidation state form +2 to +3 is easiest?

Cr

Co

Fe

Mn

Solution

straight E subscript Cr to the power of 3 plus end exponent divided by Cr to the power of 2 plus end exponent end subscript superscript straight o space space equals space minus 0.41 space straight V straight E subscript Mn to the power of 3 plus end exponent divided by Mn to the power of 2 plus end exponent end subscript superscript straight o space equals space minus space plus 1.57 space straight V straight E subscript Fe to the power of 3 plus end exponent divided by Fe to the power of 2 plus end exponent end subscript superscript straight o space equals space minus space 0.77 space straight V straight E subscript Co to the power of 3 plus end exponent divided by space Co to the power of 2 plus end exponent end subscript superscript straight o space equals space minus 1.97 space straight V

More negative value of

straight E subscript red superscript straight o

indicates better reducing agent thus easily oxidised. Thus, oxidation of Cr²⁺ to Cr³⁺ is the easiest

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