In the side reaction 2 H2O2 → 2 H2O + O2, what, respectively, are the oxidation numbers of the O atoms? Answer?
In the side reaction 2 H2O2 → 2 H2O + O2, what, respectively, are the oxidation numbers of the O atoms? Answer –1, 0, –2 0, 0, 0 –1, –2, 0 0, –1, –2
In the side reaction 2 H2O2 → 2 H2O + O2, what, respectively, are the oxidation numbers of the O atoms?
in H2O2 is-1 in H2O is -2 in O2 is 0
Consider the following reaction 2H2+O2 gives 2H2O. What gram of H2 is formed if 2G of H2 reacts with 1G of O2?
moles of O2 are :- 1gm/16moles of H2 are :- 2gm/1Now, according to the reaction,1 mole of O2 gives 2 mole of H2OTherefore, 1/16 moles will give (1/16)x2 moles.Taking H2 case,2 moles of H2 will give 2 moles of H20,But 1 mole per result mole is more than what O2 has which is (1/16)th of a mole.Thus, O2 will stop the reaction long before H2 finished and thus, becoming the limiting reagent.Thus, the mass of H20 = (1/16)*2*18Where 18 is the molar mass of H2O…..
How can I balance this half reaction if it is in a basic solution, H2O2(aq) →O2(g)?
I am not sure if this is correct, but I would start with the skeleton as you did, then add H+ to balance for H atoms, then neutralise with enough OH- (as would occur in an alkaline solution):H2O2(aq) —-> O2(g)H2O2(aq) ——-> O2(g) + 2H+(aq)H2O2(aq) ——→ O2(g) + 2H+(aq) + 2OH-(aq)H2O2(aq) + 2OH-(aq) ——→ O2(g) + 2H+(aq) + 2OH-(aq)H2O2(aq) + 2OH-(aq) ——→ O2(g) + 2H20(l) + 2e-
Tough redox reaction question?
the half reactions of the 1st one would be H2O2 (aq) --> O2 (g) + 2e + 2H+ (aq) ------------(1) oxidation ClO2 (aq) + 2e ---> ClO2- (aq) ----(2)reduction by adding (1) + (2), you get the overall reaction, which would be, H2O2 (aq) + ClO2 (aq) ---> ClO2- (aq) +O2 (g) + 2H+ (aq) likewise, for the 2nd one the half reactions would be Al (s) + 4H2O (aq) -- > Al(OH)4- (aq) + 3e + 4H+ (aq) ---(1) oxidation MnO4- (aq) + 3e + 4H+ (aq) ---> MnO2 (s) + 2H2O (aq)---(2) reduction Adding up the two you get Al (s) + MnO4- (aq) + 2H2O (aq) ---> MnO2 (s) + Al(OH)4- (aq) for the 3rd one Cl2 (g) + 2H2O (aq) ---> 2ClO- (aq) + 2e + 4H+ (aq)--- (1) oxidation Cl2 (g) + 2e ---> 2Cl- (aq)--- (2) reduction Add the two (1) + (2) to get the overall reaction 2Cl2 (g) + 2H2O (aq) ---> 2Cl- (aq) + 2ClO- (aq) + 2H+ (aq)
Balance the following reaction in basic solution?
There is no O2^2- ion by itself in solution. Do you mean peroxide, like H2O2? H2O2 does not dissociate into ions. Nor is there an oxide ion, O^2-, in aqueous solution either. Oxygen exists in solution as the hydroxide ion. You might think that something like sodium peroxide, Na2O2, might give you a peroxide ion in solution, but in fact, sodium peroxide will react with water to produce a solution of sodium hydroxide and hydrogen peroxide. Na2O2(s) + H2O(l) --> NaOH(aq) + H2O2(aq) This is the way to write the net ionic equation for the reaction, which occurs in acidic solution: Cr3+ + H2O2(aq) --> CrO4^2- + H2O(l) 2(Cr3+ + 4H2O --> CrO4^2- + 3e- + 8H+) 3(2H+ + H2O2 + 2e- --> 2H2O) ---------------------------- ------------------------------ 2Cr3+ + 3H2O2 + 6H+ + 8H2O --> 2CrO4^2- + 6H2O + 16H+ simplify 2Cr3+ + 3H2O2 + 2H2O --> 2CrO4^2- + 10H+ Whoops. You said "basic solution". Simply add enough OH- ions to both sides to "neutralize" the H+ ions. 10OH- + 2Cr3+ + 3H2O2 + 2H2O --> 2CrO4^2- + 10H+ + 10OH- 10OH- + 2Cr3+ + 3H2O2 + 2H2O --> 2CrO4^2- + 10H2O Simplify 10OH- + 2Cr3+ + 3H2O2 --> 2CrO4^2- + 8H2O
Balancing redox reaction in a basic solution?
" Chlorine " in [ Cl2O7 ] is reduced from Oxidation No. = +7 to Oxidation No. = ( +3 ) in [ ClO2- ] { Oxidation Number of Atoms in [ Cl2O7 ] : Cl = ( +7 ) ; O = ( -2 ) ; Sum = ( +7 x 2 ) + ( -2 x 7 ) = ( 0 ) } { Oxidation Number of Atoms in [ ClO2- ] : Cl = ( +3 ) ; O = ( -2 ) ; Sum = ( +3 ) + ( -2 x 2 ) = ( -1 ) = Charge of Ion } " Oxygen " in H2O2 is oxidised from Oxidation No. = ( -1 ) to Oxidation No. = ( 0 ) in O2 { Oxidation Number of Atoms in H2O2 : H = ( +1 ) ; O = ( -1 ) ; Sum = ( +1 x2 ) + ( -1 x 2 ) = ( 0 ) } { Oxidation Number of Atoms in O2 : O = ( 0 ) ; By Definition } Balancing the equation : Reduction Side : [ Cl2O7 ] + 8 e- ----> 2 [ ClO2- ] ( R Side ) ( Focus on Oxidation No. of Cl in [ Cl2O7 ] and [ ClO2- ] only ; Ignor the fact that atom and charge may not be balanced ) Oxidation Side : H2O2 ----> O2 + 2 e- ( O Side ) ( Focus on Oxidation No. of O in H2O2 and O2 only ; Ignor the fact that atom and charge may not be balanced ) 1 x ( R Side ) + 4 x ( O Side ) [[ to compensate electron ]] Cl2O7 + H2O2 ----> 2 [ ClO2- ] + 4 O2 The Redox Part ( Electron Transfer ) already balanced General Equation Balancing : ( Atoms and Positive / Negative Charges ) Add [ OH- ] on left side - - ( To enable the fact that there are negative charges on right side ) and Add [ H2O ] on right side - - ( To enable balancing of H and O on both sides ) Cl2O7 + H2O2 + 2 [ OH- ] ----> 2 [ ClO2- ] + H2O + 4 O2 Balance H2O2 , [ OH- ] on Left Side -- vs -- H2O on Right Side ( but do not change Cl2O7 , 2 [ ClO2- ] , 4 O2 ; since they are already balanced at Redox Part ) Cl2O7 + 4 H2O2 + 2 [ OH- ] ----> 2 [ ClO2- ] + 5 H2O + 4 O2 { Balanced Net Ionic Equation }