Balance the following redox equation using thesmallest integers possible ( basic

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Fe(OH)2 =>   Fe2O3 Balance Fe 2Fe(OH)2 => Fe2O3 Fe goes from 2+   to 3+. 2 Fe on both sides, so 2*1= 2 electrons 2Fe(OH)2 + => Fe2O3 + 2e- Balance O's with OH- and H's with H2O 2Fe(OH)2 +   2OH- => Fe2O3 +3H2O + 2e- ================================= reduction half reaction: CrO42-   => Cr(OH)4- Cr goes from +6 to +4 CrO42- + 2e- => Cr(OH)4- CrO42- + 2e- + 4H2O => Cr(OH)4-   + 4OH- Add bold equations          2Fe(OH)2 + 2OH- => Fe2O3 + 3H2O+ 2e-     CrO42- + 2e- +4H2O => Cr(OH)4-   + 4OH- 2Fe(OH)2 + H2O + CrO42- => Fe2O3 +Cr(OH)4- + 2OH-

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