Q.

The reversible reduction potential of pure water is –0.414V under 1 atm  H2  pressure if the reduction is considered to be 2H++2eH2 . Then calculate the [H+]  in pure water.
Write your answer in scientific notation and take it as a×10b .
Mark the value of b on the OMR (Use 2.303RT/F = 0.0591)
 

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answer is 7.

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Detailed Solution

EH+,H2=EH+,H200.05912log1[H+]2 0.414=0+0.0591  log[H+] log[H+]=6.99  ;       [H+]=1.02×107M

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