I think you are missing an S species in the first.pdf

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I think you are missing an S species in the first equation. Are you trying to from H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion [SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-. S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 + 2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is +5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 + 2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons) Solution I think you are missing an S species in the first equation. Are you trying to from H2SO4 and not H2O on the right side? 1) I2 + H2SO3 +H2O => 2HI + H2SO4 I2 is neutral; IN HI, I becomes I- (I is more electronegative than H); I is reduced S starts as the sulfite ion [SO3]2-. O has -2 charge. so S_charge + 3*(-2) = -2. S_charge = +4. In H2SO4, S is [SO4]2-. S_charge + 4*(-2) = -2; S_charge = +6 Sulfur is oxidized 2) C + 4HNO3 -> CO2 + 4NO2 + 2H2O C is oxidized from 0 to +4 in CO2 N is reduced from +5 in HNO3 to +4 in NO2 (use the same method as previous; in HNO3; [NO3]- because H is +1; N_charge + 3*-2 = -1; N_charge is +5. for NO2; N_charge + 2*-2 = 0 (neutral); N_charge = +4 3) 2FeCl3 + H2S => 2FeCl2 + 2HCl + S S exists as -2 in H2S, and becomes 0 in S; it is oxidized (it has lost 2 electrons) Fe exists as +3 in FeCl3 (Cl as Cl-) and becomes +2 in FeCl2; it is reduced (gains 1 electrons).

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