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Calculate the reduction potential for the following half cells at 25^@C Pt|Fe^(2+) (0.1M)-Fe^(3+) (0.01M), E_(Fe^(3+),Fe^(2+))^@=+0.77V |
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Answer» Solution :`FE^(3+)+e LEFTRIGHTARROW Fe^(2+)` (reduction) `E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^(2+))-0.0591/n LOG""([Fe^(2+)])/([Fe^(3+)])` `=0.77- 0.0591/1 log"" 0.1/0.01=0.7109V` |
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