Question | Answer |
ZINC HALF EQUATION ... | ZN(S) ZN2+ + 2E-+ |
OXIDATION REACTION ELECTRON ON THE RIGHT HAND SIDE. | + CHARGES MUST BALANCE + STATE SYMBOLS ARE INCLUDED. +THE REACTION SHOULD BE ON LEFT. |
REDUCTION HALF EQUATIONS ... | 2H (AQ) + 2EH2(G) ELECTRONS ON THE LEFT HAND SIDE. |
WRITE AN IONIC HALF EQUATION FOR OXIDATION OF FE2+... | FE 2+ IONS CAN LOOSE ELECTRON = FE3+ SO FE2+(AQ) FE3+ (AQ) + E- |
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). |
SOME OXIDISING AGENTS REQUIRE PRESENCE OF ... | ACID E.G. SULPHURIC ACID, KMG(V11)IS A STRONG OXIDISING AGENT. |
THE MANGANESE IN MNO4-IONIS REDUCED TO... | MN2+IONS |
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 |
IN ACID SOLUTION H2O2 HYDROGEN PEROXIDE ACTS AS ... | OXIDISING AGENT REDUCED TO WATER THE EQUATIONS HAS |
H+ AND ELECTRONS ON THE LEFT H2O2(AQ) + 2H+(AQ) +2E- 2H2O(L). | |
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. |
THERE ARE 7 OXYGEN ATOMS IN CR2O72- SO ... | 7 WATER MOLECULES PRODUCED 14 HYDROGEN IONS ON LEFT. |
AND SOLVE FOR X TO GET CHARGES. | |
THE ELECTRONS ON LEFT FOR ... | REDUCTION, |
ON RIGHT FOR .. | OXIDATION. + |
IF REACTION TAKES PLACE IN ACID ADD... | H+ TO RIGHT AND WATER ON LEFT. + MAKE SURE EQUATIONS BALANCES FOR ATOMS AND CHARGES. |
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. |
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 |
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. |
OXIDATION NUMBER OF MG IN MGO .... | IS 2+. ALSO CL = -1 OXIDATION NUMBER. |
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. |
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 , |
BUT THE SULPHUR MORE | ELECTRONEGATIVE IS ASSIGNED 2 ELECTRONS GAINING CHARGE OF 2-=OXIDATION NUMBER IS NOT IONIC THOUGH. |
DEDUCTION OF OXIDATION NUMBERS OXIDATION NUMBER OF UNCOMBINED ELEMENT IS | = 0 E.G. CL2 |
THE OXIDATION NUMBER IN MONATOMIC ELEMENT IS | THE CHARGE E.G.CU2+ ION IN THE 2+ STATE. |
ALL GROUP ONE AND 2 METALS HAVE AN OXIDATION NUMBER OF . | +1 +2 IN THEIR COMPOUNDS * FLUORINE HAS OXIDATION NUMBER -1 IN ITS COMPOUND. |
HYDROGEN OXIDATION WHEN COMBINED WITH METAL IS... | -1. |