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Electrochemistry

Question
CBSEENCH12010129

Calculate emf of the following cell at 298 K: Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given E0 cell = +2.71 V, 1 F = 96500 C mol-1]

Solution

The cell reaction can be represented as:

Mg(s) + Cu2+(aq.) ---> Mg+(aq.) + Cu(s)

 Given:

  straight E subscript cell superscript 0 =+2.71 V

T = 298 K

According to the Nernst equation:

  straight E equals space straight E subscript cell superscript 0 space minus fraction numerator 0.0591 over denominator 2 end fraction log space fraction numerator left square bracket Mg to the power of 2 plus end exponent right square bracket over denominator left square bracket Cu to the power of 2 plus end exponent right square bracket end fraction space equals 2.71 minus fraction numerator 0.0591 over denominator 2 end fraction log fraction numerator 0.1 over denominator 0.01 end fraction

2.71 minus 0.295 space log space 10 equals space 2.71 minus 0.0295 space equals 2.6805 space straight V space

 

 

=2.71-0.0295 log 10 = 2.71-0.0295

 

=2.6805 V