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The electrochemical cell described by the cell notation has an Eo of -0.37 V. Calculate the maximum electrical work (kJ) the cell has done if 331.73 g of Cu(s) (Molar Mass - 63.55 g/mol) reacts. Round your answer to 3 significant figures.

Cu(s) l Cu+(aq) ll Sn4+(aq), Sn2+(aq) l Pt(s)

St. Red. Pot. (V)
Cu+/Cu +0.52
Sn4+/Sn2+ +0.15
Faraday's Constant
F = 96485 C

convert 331g to mole= 5.21
G=-nFE=2(5.21)(96485)(-.37)=372kJ, why do i have to divide it in half to get 186kJ

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