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Use the following reduction potentials to answer questions 208)+2e 02(g) + 4H(aq) +4e- Br20 +2e 2) +2e Niat) + 2e- 2H20u +2e 4-7 Eoed 1.87V Ered = 1.40V Eoed 1.09v Ered = 0.54 V e0.23v Ered =-0.41 V ed0.73V Ere,--3.04V 2H2O(1) ー→ 21(aq) Ni(s) --+ ー→ (aq) aq → Cr Litaa) e ー→ Li(s)
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Answer #1

一→ @냐 ED 足 Oxidation →(Anodic geaclion (al anode) Td total sceaction anode Reduclion reactions

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