Calculate E for the cell below using the correct half-reactions Fe +2 + 2e- ----- Fe EO = -0.440 Fe +3 +1e- Would this cell be spontaneous as written? Zn +2 (aq) + 2Fe+2 (aq) ------ 2Fe+ (aq) + Zn (s) +3 Cell potential to two decimal places: Spontaneous (yes or no): ----- Fe +2 EO = +0.771 Zn +2 +2e-- Zn E = -0.763-.

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Calculate E for the cell below using the correct half-reactions Fe +2 + 2e- -----> Fe EO = -0.440 Fe +3 +1e- Would this cell be spontaneous as written? Zn +2 (aq) + 2Fe+2 (aq) ------> 2Fe+ (aq) + Zn (s) +3 Cell potential to two decimal places: Spontaneous (yes or no): -----> Fe +2 EO = +0.771 Zn +2 +2e--> Zn E = -0.763-.

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