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Electrochemistry

Question
CBSEENCH12010780
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Given:  straight E subscript Fe to the power of 3 plus end exponent divided by Fe end subscript superscript 0 space equals space minus 0.036 straight V comma space straight E subscript Fe to the power of 2 plus end exponent divided by Fe end subscript superscript 0 space equals space minus 0.439 straight V The value of standard electrode potential for the change Fe subscript left parenthesis aq right parenthesis end subscript superscript 3 plus end superscript space plus space straight e to the power of minus space rightwards arrow Fe to the power of 2 plus end exponent space left parenthesis aq right parenthesis will be

  • -0.072

  • 0.385 V

  • 0.770 V

  • -0.270 V

Solution

C.

0.770 V

because space Fe to the power of 3 plus end exponent space plus space 3 straight e to the power of minus space rightwards arrow space Fe space semicolon space straight E to the power of 0 space equals space minus space 0.036 space straight V therefore space increment straight G subscript 1 superscript 0 space equals space minus space nFE to the power of 0 space equals space minus space 3 straight F space left parenthesis negative 0.036 right parenthesis space equals space plus space 0.108 space straight F Also space Fe to the power of 2 plus end exponent space plus space 2 straight e to the power of minus space rightwards arrow space Fe semicolon space straight E to the power of 0 space equals space minus space 0.439 space straight V therefore space increment straight G subscript 2 superscript 0 space equals space minus nFE to the power of straight o space equals space 2 straight F space left parenthesis negative 0.439 right parenthesis space equals space 0.878 straight F To space find space straight E to the power of straight o space for space Fe subscript left parenthesis aq right parenthesis end subscript superscript 3 plus end superscript space plus straight e to the power of minus space rightwards arrow space Fe to the power of 2 plus end exponent space left parenthesis aq right parenthesis increment straight G to the power of straight o space equals space minus nFE to the power of straight o space equals space minus space 1 FM to the power of straight o because space straight G to the power of straight o space equals space straight G subscript 1 superscript 0 minus straight G subscript 2 superscript 0 straight G to the power of 0 space equals space straight G subscript 1 superscript straight o space minus straight G subscript 2 superscript straight o therefore space straight G to the power of straight o space equals space 0.108 space straight F space minus space 0.878 space straight F therefore space minus FE to the power of straight o space equals space plus 0.108 space straight F space minus space 0.878 straight F therefore space straight E to the power of 0 space equals space 0.878 space minus 0.108 space equals space 0.77 space straight v

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