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Question

ntGiven electrode potentials:n ntFe3+ plus e ---------Fe2+ ; Eo=0.771n ntI2+2e ----------------2I- ; Eo =0.536n ntEo cell for the cell reactionn nt2Fe3+ plus 2I- ----------2Fe2+ plus I2 is -n nt1. (2x0.771-0.536) =1.006 voltsn nt2. (0.771-0.5x0.536)=0.503 voltsn nt3. 0.771-0.536=0.253 voltsn nt4. 0.536-0.771=-0.235 voltsn

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Solution

99996126377-0-0

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