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Calculate the reduction potential for the following half cells at 25^@C Cl_2|Cl^(-)(2 times 10^-5 M), E_(Cl_2,Cl^-)^@=+1.36V |
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Answer» SOLUTION :`1/2 Cl_2+e leftrightarrow Cl^-` (reduction) `E_(Cl_2,Cl^-)=E_(Cl_2,Cl^-)^@-0.0591/n LOG [Cl^-]` [ See example 1] `=1.36 -0.0591/1 log (2 times 10^-5)=1.6377 V` |
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