What is the Oxidation-Reduction reaction of KMnO4 + H2O2 + H20 + H2SO4?
KMnO4 is the oxidising agent, and H2O2 is the reducing agent. Mn is reduced from the +7 state to the +2 state under acidic conditions, and the O from H2O2 is oxidised from -1 to 0 and will form O2 gas. If you need to balance it, split into half reactions and ignore spectator ions (such as K and SO4^2-). Balance non-redox O by adding H2O, H by adding H^+, and electrons by checking charges and adding e^- to the more positive side so charges balance. Add the half-reactions so the electrons cancel, and finally, add in the remaining ions. MnO4^- → Mn^2+ MnO4^- → Mn^2+ + 4H2O (balance O) MnO4^- + 8H^+ → Mn^2+ + 4H2O (balance H) MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O (balance electrons; both sides have a total charge of +2) H2O2 → O2 H2O2 → O2 + 2H^+ H2O2 → O2 + 2H^+ + 2e^- You need 2 times the Mn half reaction, and 5 times the O half reaction to balance electrons. 2MnO4^- + 16H^+ + 10e^- → 2Mn^2+ + 8H2O 5H2O2 → 5O2 + 10H^+ + 10e^- ————————————————————— 2MnO4^- + 5H2O2 + 16H^+ + 10e^- → 2Mn^2+ + 5O2 + 8H2O + 10H^+ + 10e^- Now eliminate common entities, such as electrons and excess H^+ ions. 2MnO4^- + 5H2O2 + 6H^+ → 2Mn^2+ + 5O2 + 8H2O This is the net ionic equation. For the complete equation, add in the missing ions for the reactants first (these were given in the problem, so just restore the original reagents), add in the extra ions to the product side, then work out what the remaining products will be by combining ions appropriately. 2KMnO4 + 5H2O2 + 3H2SO4 → 2Mn^2+ + 5O2 + 2K^+ + 3SO4^2- + 8H2O 2KMnO4 + 5H2O2 + 3H2SO4 → MnSO4 + 5O2 + K2SO4 + 8H2O
What would be the oxidation and reduction half reactions for the equation Cu+H2(SO4)+H2O2----->Cu(SO4)+O...
You wrote: Cu(s) + H2SO4(aq) + H2O2(aq) --> CuSO4(aq) + O2(g) + H2O(l) but ... Cu(s) --> Cu2+ + 2e- H2O2 --> O2 + 2H+ + 2e- See a problem? Two oxidation reactions, and no reduction. Oxygen will not be produced in this reaction, only copper(II) sulfate and water. Cu(s) + H2SO4(aq) + H2O2(aq) --> CuSO4(aq) + H2O(l) Cu(s) --> Cu2+ + 2e- 2H+ + H2O2 + 2e- --> 2H2O(l) ------------------------ ------------------------- Cu(s) + H2O2 + 2H+ --> Cu2+ + 2H2O The equation directly above is the net ionic equation, and the one below in the molecular equation. Cu(s) + H2SO4(aq) + H2O2(aq) --> CuSO4(aq) + 2H2O(l)
What is the reaction between oxalic acid and potassium permanganate?
This is a redox reaction. KMnO4 oxidises (COOH)2 to give you CO2 and H2O.MnO4- + 8H+ + 5e- →Mn2+ + 4H2O(C2O4)2- → 2CO2 + 2e-Balanced:2MnO4- + 16H+ + 5(C2O4)2- → 10CO2 + 2 Mn2+ + 8 H2OSide note:This reaction is an autocatalytic reaction where Mn2+ catalyses the reaction. The start of the reaction is relatively slow due to the absence of the catalyst Mn2+ so usually the oxalic acid solution is warmed to about 65–70oC before a titration with KMnO4 is done. Alternatively, a few drops of MnSO4 can be added into the conical flask before the start of the titration as well.
What happens when hydrogen peroxide reacts with acidified KMnO4?
You get one of those occasions where hydrogen peroxide acts as a reducing agent rather than its more usual role as an oxidant. It’s oxidised to oxygen gas, whilst the permanganate is reduced to manganese dioxide:2 KMnO4 + 3 H2O2 → 2 MnO2 + 2 KOH + 2 H2O + 3 O2
Pla help balance this chem equation:FeC2O4 + KMnO4 + H2SO4 -> Fe2(SO4)3 +CO2 + MnSO4 + K2SO4 + H2O?
Reduction : ( R-Half ) Oxidation No. of Mn changes from +7 to +2 [ MnO4- ] + 5 [ e- ] ---- > [ Mn++ ] Oxidation : ( O-Half ) [ Fe++ ] ----> [ Fe+++ ] + 1 [ e- ] Oxidation No. of Fe changes from +2 to +3 [ C2O4-- ] ---- > 2 CO2 + 2 [ e- ] Oxidation No. of C x 2 atoms changes from +3 x 2 to +4 x 2 { ( R-Half ) x 3 } + { ( R-Half ) x 5 } ----- To compensate electrons 3 [ MnO4- ] + 5 [ Fe++ ] + 5 [ C2O4-- ] ----> 3 [ Mn++ ] 5 [ Fe+++ ] + 10 CO2 ( Redox Part already Balanced ) General Balancing of Equation : ( Atoms , Positive Charges , Negative Charges ) Add [ H+ ] on Left Side ; Add [ H2O ] on Right Side Balancing of O and H in [ MnO4- ] , [ H+ ] , [ H2O ] ; but do not change the amount of components already balanced in Redox Part ) 3 [ MnO4- ] + 5 [ Fe++ ] + 5 [ C2O4-- ] + 24 [ H+ ] ----> 3 [ Mn++ ] 5 [ Fe+++ ] + 10 CO2 + 12 H2O ( Balanced - Net Ionic Equation ) For Balanced - Full Equation : Multiply both sides by 2 Add [ SO4-- ] on both sides Add [ K+ ] on both sides 6 KMnO4 + 10 FeC2O4 + 24 H2SO4 ----> 3 K2SO4 + 6 MnSO4 + 5 Fe2(SO4)3 + 20 CO2 + 24 H2O ( Balanced - Full Equation )
Balancing Redox Reactions Run in Acidic | Base Conditions Using the Oxidation Number Method?
I would be much thankful if anyone could teach me how to balance redox reactions run in acidic or basic conditions using the “oxidation number” method. So far I've no problems with the half-reaction technique. KMnO4 + H2SO4 + H2O2 - - > KHSO4 + MnSO4 + H2O + O2 FeSO4 + H2SO4 + H2O2 - - > Fe2(SO4)3 + H2O