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OXIDATION NO

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Answer
ZINC HALF EQUATION ...   ZN(S) ZN2+ + 2E-+  
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OXIDATION REACTION ELECTRON ON THE RIGHT HAND SIDE.   + CHARGES MUST BALANCE + STATE SYMBOLS ARE INCLUDED. +THE REACTION SHOULD BE ON LEFT.  
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REDUCTION HALF EQUATIONS ...   2H (AQ) + 2EH2(G) ELECTRONS ON THE LEFT HAND SIDE.  
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WRITE AN IONIC HALF EQUATION FOR OXIDATION OF FE2+...   FE 2+ IONS CAN LOOSE ELECTRON = FE3+ SO  FE2+(AQ) FE3+ (AQ) + E-  
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WRITE IONIC HALF EQUATION FOR REDUCTION OF CHLORINE...   CHLORINE ATOMS GAIN 2 ELECTRON AND ARE REDUCED TO CHLORIDE ION. CL2(G) + 2E-  2CL-(AQ) AND CAN BE HALVED TO 1/2CL2(G) + E- CL-(AQ).  
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SOME OXIDISING AGENTS REQUIRE PRESENCE OF ...   ACID E.G. SULPHURIC ACID, KMG(V11)IS A STRONG OXIDISING AGENT.  
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THE MANGANESE IN MNO4-IONIS REDUCED TO...   MN2+IONS  
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HALF EQUATION FOR MAGANESE...   IS MN04-(AQ) + 8H+(AQ) + XE-  MN2+ (AQ) + 4H20(L). CHARGE ON RIGHT IS 2+ BUT CHARGE ON LEFT IS -1 + 8-X = 7 – X (7-X=+2) X = 5  
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IN ACID SOLUTION H2O2 HYDROGEN PEROXIDE ACTS AS ...   OXIDISING AGENT REDUCED TO WATER THE EQUATIONS HAS  
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H+ AND ELECTRONS ON THE LEFT H2O2(AQ) + 2H+(AQ) +2E- 2H2O(L).    
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WRITE IONIC HALF EQUATION FOR DICHROMATE (VI) IONS TO CR3+ IONS IN THE ACID SOLUTION...   ELECTRONS ON LEFT BECAUSE REDUCTION. THE SOLUTION IS ACIDIC SO H+ IONS ON LEFT.  
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THERE ARE 7 OXYGEN ATOMS IN CR2O72- SO ...   7 WATER MOLECULES PRODUCED 14 HYDROGEN IONS ON LEFT.  
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AND SOLVE FOR X TO GET CHARGES.    
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THE ELECTRONS ON LEFT FOR ...   REDUCTION,  
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ON RIGHT FOR ..   OXIDATION. +  
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IF REACTION TAKES PLACE IN ACID ADD...   H+ TO RIGHT AND WATER ON LEFT. + MAKE SURE EQUATIONS BALANCES FOR ATOMS AND CHARGES.  
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OVERALL HALF EQUATIONS YOU MUST ...   + MULTIPLY ONE OR BOTH EQUATIONS BY NUMBER SO NUMBER OF ELECTRONS THE SAME ON BOTH SIDES. + STEP 2 : ADD THE HALF EQUATIONS TOGETHER. +STEP 3 CHECK THAT BOTH REACTS IN AT ARE ON LEFT SIDE. THE ELECTRONS ON THE LEFT HAND CANCEL ELECTRON ON RIGHT.  
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REACTION BETWEEN AQUEOUS SILVER NITRATE AND COPPER METAL IS ...   A REDOX REACTION. SILVER IONS ARE REDUCED TO SILVER ATOMS AND COPPER ARE OXIDISED TO COPPER IONS  
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CAN DETERMINE THE OVERALL EQUATION BY ...   ( 2HNO3 + H2S  2NO2 + S +2H20. FIRST WORK OUT WHICH HAS BEEN OXIDISED AND WRITE ELECTRONS ON RIGHT HAND SIDE. H2S  2H+ + S + 2E- [ 2HNO3 + 2H+ + 2E-  2NO2 + 2H2O]. NITRIC ACID IS AN OXIDISING AGENT.  
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OXIDATION NUMBER OF MG IN MGO ....   IS 2+. ALSO CL = -1 OXIDATION NUMBER.  
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THE OXIDATION NUMBER OF AN ELEMENT IN A COMPOUND OR ION IS ...   THE CHARGE THAT THE ELEMENT WOULD HAVE IF THE COMPOUND WERE FULLY IONIC.  
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OXIDATION NUMBER CALCULATE ON THE FACT THAT   THE BONDING ELECTRONS ARE ASSIGNED TO THE MORE ELECTRON NEGATIVE ATOM... E.G. H2S THE BONDING IS COVALENT AND THE SULFATE ION SHARES PAIR OF ELECTRONS WITH THE H IONS ,  
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BUT THE SULPHUR MORE   ELECTRONEGATIVE IS ASSIGNED 2 ELECTRONS GAINING CHARGE OF 2-=OXIDATION NUMBER IS NOT IONIC THOUGH.  
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DEDUCTION OF OXIDATION NUMBERS  OXIDATION NUMBER OF UNCOMBINED ELEMENT IS   = 0 E.G. CL2  
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THE OXIDATION NUMBER IN MONATOMIC ELEMENT IS   THE CHARGE E.G.CU2+ ION IN THE 2+ STATE.  
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ALL GROUP ONE AND 2 METALS HAVE AN OXIDATION NUMBER OF .   +1 +2 IN THEIR COMPOUNDS * FLUORINE HAS OXIDATION NUMBER -1 IN ITS COMPOUND.  
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HYDROGEN OXIDATION WHEN COMBINED WITH METAL IS...   -1.  
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