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Electrochemistry

Question
CBSEENCH12011397
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If  straight E to the power of straight o subscript Fe to the power of 2 plus end exponent divided by Fe end subscript space equals space minus space 0.441 space straight V space and space straight E to the power of straight o subscript Fe to the power of 3 plus end exponent divided by Fe to the power of 2 plus end exponent end subscript space equals space 0.771 space space straight V comma the standard emf of the reaction:

Fe + 2 Fe3+ →3Fe2+

will be:

  • 0.330 V

  • 1.653 V

  • 1.212 V

  • 0.111 V

Solution

C.

1.212 V

Given that,
straight E subscript Fe to the power of 2 plus divided by Fe end exponent end subscript superscript 0 space equals space minus 0.441 space straight V space So comma space Fe space rightwards arrow Fe to the power of 2 plus end exponent space plus space 2 straight e space to the power of minus comma space straight E to the power of straight o space space equals space plus 0.441 space straight V space... space left parenthesis straight i right parenthesis and space straight E subscript Fe to the power of 3 plus divided by end exponent Fe to the power of 2 plus end exponent end subscript superscript 0 space equals space 0.771 space straight V So comma space Fe to the power of 3 plus end exponent space plus straight e to the power of minus space rightwards arrow Fe to the power of 2 plus end exponent comma space straight E to the power of straight o space equals space 0.771 space straight V space.... left parenthesis ii right parenthesis Cell space reaction left parenthesis straight i right parenthesis space Fe space rightwards arrow Fe to the power of 2 plus end exponent space plus space 2 straight e to the power of minus comma space space space space space space space space space space space space space space space straight E to the power of straight o space equals space 0.441 space straight V left parenthesis ii right parenthesis 2 Fe to the power of 3 plus end exponent space plus space 2 straight e to the power of minus space rightwards arrow space 2 Fe to the power of 2 comma plus end exponent space space space space space space space straight E to the power of straight o space equals space plus 0.771 space straight V _________________________________________ Fe to the power of 2 plus end exponent space plus space 2 Fe to the power of 3 plus end exponent space rightwards arrow 3 Fe to the power of 2 plus end exponent comma space space space space space space space space space straight E subscript cell superscript straight o space equals space 1.212 space straight V
So, on the basis of cell reaction following half-cell reactions are written
At anode:
Fe → Fe2+ + 2e-       (oxidation)
At cathode:
2Fe3+ + 2e- →2Fe2+ (reduction)
So,
straight E subscript cell superscript 0 space equals space straight E subscript cathode superscript straight o minus space straight E subscript anode superscript 0 space equals space straight E to the power of 0 subscript Fe to the power of 3 plus end exponent divided by Fe to the power of 2 plus end exponent end subscript space minus space straight E to the power of 0 subscript Fe to the power of 2 plus end exponent divided by Fe end subscript left parenthesis 0 plus.771 right parenthesis minus left parenthesis negative 0.441 right parenthesis space equals space space plus 1.212 space straight V

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