Redox reactions question?
You are right that the answer is (1). Why? Because Cl is more reactive than Br and will replace it in the compound NaBr. In all the other choices, the halogen in the compound is more reactive than the halogen trying to replace it so there is no reaction. In terms of reactivity, F > Cl > Br > I.
HOW TO PICK OUT REDOX REACTIONS?
You have to assign oxidation numbers to the atoms in the equation and see if any of those oxidation numbers are changing. Look at your book for the detailed rules for assigning oxidation numbers. In the first equation, by element: P4 oxidation # = 0 HClO H = +1 O = -2 Cl = +1 H2O H = +1 O = -2 H3PO4 H = +1 O = -2 P = +7 HCl H = +1 Cl = -1 So, in that reaction the oxidation numbers of P and of Cl are changing. One is being oxidized (P) and the other is being reduced (Cl) Do the same for all of your reactions, and you 'll see that 3 of them are Redox reactions. Only one is not.
Oxidation-reduction reactions?
Which of the following are redox reactions? For those that are, indicate which element is oxidized and which is reduced. For those that are not, indicate whether they are precipitation or neutralization reactions. a.) Pb(s) + 10HClO(aq) +6H2O(l) -> 4H3PO4(aq) + 10HCL b.) Br2(l) + 2K -> 2KBr(s) c.) CH3CH2OH(l) + 3O2 -> 3H2O(l) +CO2(g) d.) ZnCl2(aq) 2NAOH(aq) -> Zn(OH)2(s) +2NaCl(aq) Please help me with this question. The best answer gets a nice compliment ;)
F2 (g) + 2KI (aq) -> 2KF (aq) + I2 (aq) Oxidation?
Net Ionic Equation: F2 + I- --> F- + I2 K is a spectator ion. Assign oxidation numbers to determine what is oxidized and what is reduced. F2 = 0 I- = -1 F- = -1 I2 = 0 The one that gains electrons (becomes more negative) is reduced and the one that loses electrons (becomes more positive) is oxidized. Since F went from 0 to -1, it was reduced. Since I went from -1 to 0, it was oxidized.
Which of the following is NOT a redox reaction?
a. ) 2 KI(aq) + Br2(aq) ---> 2 KBr(aq) + I2(s); >>> in this one, iodide from the left is oxidized to iodine of the right; bromine molecule on the left is being reduced to bromide on the left. NOT this one! b. ) 2 F2(g) + 2 H2O(l) ---> 4 HF(aq) + O2(g); >>> in this one, oxygen from water on the left is oxidized to oxygen molecule on the right; fluorine molecule on the left is being reduced to fluoride on the left. NOT this one! c. ) Cu(s) + 2 HNO3(aq) ---> Cu(NO3)2(aq) + H2(g); >>> in this one, copper metal from the left is oxidized to copper (II)of the right; hydrogen ion on the left is being reduced to hydrogen on the left. NOT this one! d. ) H2CO3(aq) --> H2O(l) + CO2(g); Here, all the atoms have the same charge left and right: hydrogen is +1, carbon is +4, oxygen is -2. NO REDOX! This is your answer! Hope this helped!
HELP! Need to predict & balance Redox Reactions?
*** a *** you wrote...2KI(aq) + Br2(g) --> I2(aq) + 2KBr(aq) try this... 2KI(aq) + Br2(aq) --> I2(aq) + 2KBr(aq) bromine is a liquid at room temperature and dissolves in water. *** b *** you wrote...Mn3(SO4)2(aq) and 2AlSO4(aq) SO4 has a -2 charge. so if Mn is +2.. the salt is MnSO4... Al is +3.. SO4 -2.. therefore aluminum sulfate is Al2(SO4)3 balanced equation... 2 Al(s) + 3 MnSO4 (aq) --> Al2(SO4)3(aq) + 3 Mn(s) *** c *** PCl3(l) + Cl2(g) ---> PCl5(s) watch the phases. if you don't know.. google the melting point and boiling point of each of these... remember.. the question stated AT SATP.. standard atmospheric temperature and pressure... 25C and 100 kPa (approx.. 1 atm) *** d *** C3H8(l) + 5O2(g) --> 3CO2(g) + 4H2O(l) pentane... mp = -130 C.. bp = 36C.. liquid at SATP...not gas... *** e *** 2 KI(aq) + Br2(aq) --> I2(aq) + 2 KBr(aq)... right? overall ionic... 2 K(+1) (aq) + 2 I(-1) (aq) + Br2(aq) --> I2(aq) + 2 K(+1) (aq) + 2 Br(-1) (aq) net ionic...cancel common ions... 2 I(-1) (aq) + Br2(aq) --> I2(aq) + 2 Br(-1) (aq) *** II a *** try... 2 LiCl(l) ----> 2 Li(l) + Cl2(g) why? because electrolysis usually occurs in the liquid phase. LiCl is a liquid at temperatures.... 605 C < T < 1382 C Li is a liquid at 605 C < T < 1382 C Cl2 is a gas in that temperature range... *** II b *** SnCl2(aq) + Co(s) ----> Sn(s) + CoCl2(aq)