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Identify the oxidation states of each element in the following redox reaction:Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O.

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In this redox reaction, we can identify the oxidation states of each element as follows:1. Cu  Copper :In its elemental form, the oxidation state of Cu is 0.2. HNO3  Nitric acid :For H, the oxidation state is +1.For O, the oxidation state is -2.For N, we can determine its oxidation state using the formula: oxidation state of N = x +1  + x + 3 -2  = 0x - 6 = 0x = +5So, the oxidation state of N in HNO3 is +5.3. Cu NO3 2  Copper II  nitrate :For Cu, the oxidation state is +2  as indicated by the Roman numeral II .For N and O in the nitrate ion  NO3- , the oxidation states are the same as in HNO3: N is +5 and O is -2.4. NO2  Nitrogen dioxide :For O, the oxidation state is -2.For N, we can determine its oxidation state using the formula: oxidation state of N = xx + 2 -2  = 0x - 4 = 0x = +4So, the oxidation state of N in NO2 is +4.5. H2O  Water :For H, the oxidation state is +1.For O, the oxidation state is -2.In summary, the oxidation states in the redox reaction are:Cu: 0HNO3: H +1 , N +5 , O -2 Cu NO3 2: Cu +2 , N +5 , O -2 NO2: N +4 , O -2 H2O: H +1 , O -2

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