Asked by eng
The electrochemical cell described by the cell notation has a standard cell potential of -0.35 V. Calculate the Wmax (kJ) the cell has done if 1893.5 g of MnO42-(aq) (Molar Mass - 118.94 g/mol) forms. Round your answer to 3 significant figures.
Pt(s) | Hg22+(aq), Hg2+(aq)||MnO4-(aq), MnO42-(aq) | Pt(s)
since there are 2 transferred eletrons
then G=-nFE=(2)(.35)(96485)
Pt(s) | Hg22+(aq), Hg2+(aq)||MnO4-(aq), MnO42-(aq) | Pt(s)
since there are 2 transferred eletrons
then G=-nFE=(2)(.35)(96485)
Answers
Answered by
DrBob222
Other than omitting a negative sign on the -nEF I don't see anything wrong but you haven't answered the question.
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