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Electrochemistry

Question
CBSEENCH12011200
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The electrode potentials for 

Cu2+ (aq) + e- → Cu+ (aq) and Cu+ (aq) + e- →Cu (s)

are +0.15 V and +0.50V respectively. The value of  straight E subscript bevelled Cu to the power of 2 plus end exponent over Cu end subscript superscript straight o will be 

  • 0.325 V 

  • 0.650 V

  • 0.150 V

  • 0.500 V

Solution

A.

0.325 V 

Cu to the power of 2 plus end exponent space plus space straight e to the power of minus space Cu to the power of plus space semicolon straight E subscript 1 superscript 0 space equals space 0.15 space straight V semicolon space increment straight G subscript 1 superscript straight o space space equals space minus straight n subscript 1 straight E subscript 1 superscript straight o straight F Cu to the power of plus space plus straight e to the power of minus space space Cu semicolon straight E subscript 2 superscript 0 space equals space 0.50 straight V space semicolon space increment straight G subscript 1 superscript 0 space equals space minus space straight n subscript 2 straight E subscript 2 superscript straight o straight F ________________________________ Cu to the power of 2 plus end exponent space plus 2 straight e to the power of minus space space Cu space semicolon straight E subscript 2 superscript straight o space equals space ? semicolon space increment straight G to the power of straight o space equals space minus nE to the power of straight o straight F increment straight G to the power of straight o space equals space increment straight G subscript 1 superscript straight o space plus increment straight G subscript 2 superscript straight o Or space minus 2 straight E to the power of straight o straight F space equals space minus space 1 straight F space space straight x space 0.15 space plus space left parenthesis negative 1 straight F space straight x space 0.50 right parenthesis Or minus 2 straight E to the power of straight o straight F space equals space minus space 0.15 space straight F space minus space 0.50 space straight F Or minus 2 FE to the power of straight o space equals space minus straight F space left parenthesis 0.15 space plus space 0.50 right parenthesis straight E to the power of straight o space equals space fraction numerator 0.65 over denominator 20 end fraction space equals space 0.325 space straight V

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