Based on the oxidation state of chromium in the compounds KCrO2, K2Cr2O7, determine which of them

| August 17, 2021 | education

Based on the oxidation state of chromium in the compounds KCrO2, K2Cr2O7, determine which of them can be a reducing agent, which can be an oxidizing agent. Based on the electronic circuits, arrange the coefficients in the reaction equation: HBr + H2SO4 (conc.) = Br2 + SO2 + H2O. Indicate which substance is a reducing agent, which is an oxidizing agent; which one is reduced, which one is oxidized?

1. K2Cr2O7 – the oxidation state of chromium in this compound is +6. Chromium can only act as an oxidizing agent. KCrO2 – in this compound, chromium is oxidized to +3 (can be a reducing agent).
2.2HBr + H2SO4 = Br2 + SO2 + 2H2O
in this reaction, sulfur oxidizes bromine and is reduced from (+6) to (+4), and bromine is oxidized from (-1) lo (0). Answer: sulfur is an oxidizing agent, bromine is a reducing agent



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