balancing the equations by ion electron method
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A chemical equation is symbolic representation of a chemical reaction where the reactant entities are given on the left hand side and the product entities on the right hand side. The coefficients next to the symbols and formulae of entities are the absolute values of the stoichiometric numbers. The first chemical equation was diagrammed by Jean Beguin in 1615.
A chemical equation consists of the chemical formulas of the reactants (the starting substances) and the chemical formula of the products (substances formed in the chemical reaction). The two are separated by an arrow symbol (\rightarrow, usually read as "yields") and each individual substance's chemical formula is separated from others by a plus sign.
As an example, the formula for the burning of methane can be denoted:
- CH|4| + 2 O|2| \rightarrow CO|2| + 2 H|2|O
This equation would be read as "CH four plus O two produces CO two and H two O." But for equations involving complex chemicals, rather than reading the letter and its subscript, the chemical formulas are read using IUPAC nomenclature. Using IUPAC nomenclature, this equation would be read as "methane plus oxygen yields carbon dioxide and water."
This equation indicates that oxygen and CH4 react to form H2O and CO2. It also indicates that two oxygen molecules are required for every methane molecule and the reaction will form two water molecules and one carbon dioxide molecule for every methane and two oxygen molecules that react. The stoichiometric coefficients (the numbers in front of the chemical formulas) result from the law of conservation of mass and the law of conservation of charge (see "Balancing Chemical Equation" section below for more information).
Symbols are used to differentiate between different types of reactions. To denote the type of reaction:
- "=" symbol is used to denote a stoichiometric relation.
- "\rightarrow" symbol is used to denote a net forward reaction.
- "\rightleftarrows" symbol is used to denote a reaction in both directions.
- "\rightleftharpoons" symbol is used to denote an equilibrium.
Physical state of chemicals is also very commonly stated in parentheses after the chemical symbol, especially for ionic reactions. When stating physical state, (s) denotes a solid, (l) denotes a liquid, (g) denotes a gas and (aq) denotes an aqueous solution.
If the reaction requires energy, it is indicated above the arrow. A capital Greek letter delta (\Delta) is put on the reaction arrow to show that energy in the form of heat is added to the reaction. h\nu is used if the energy is added in the form of light.
Balancing chemical equations
The law of conservation of mass dictates the quantity of each element does not change in a chemical reaction. Thus, each side of the chemical equation must represent the same quantity of any particular element. Similarly, the charge is conserved in a chemical reaction. Therefore, the same charge must be present on both sides of the balanced equation.
One balances a chemical equation by changing the scalar number for each chemical formula. Simple chemical equations can be balanced by inspection, that is, by trial and error. Another technique involves solving a system of linear equations.
Ordinarily, balanced equations are written with smallest whole-number coefficients. If there is no coefficient before a chemical formula, the coefficient 1 is understood.
The method of inspection can be outlined as putting a coefficient of 1 in front of the most complex chemical formula and putting the other coefficients before everything else such that both sides of the arrows have the same number of each atom. If any fractional coefficient exist, multiply every coefficient with the smallest number required to make them whole, typically the denominator of the fractional coefficient for a reaction with a single fractional coefficient.
As an example, the burning of methane would be balanced by putting a coefficient of 1 before the CH4:
- 1 CH|4| + O|2| \rightarrow CO|2| + H|2|O
Since there is one carbon on each side of the arrow, the first atom (carbon) is balanced.
Looking at the next atom (hydrogen), the right hand side has two atoms, while the left hand side has four. To balance the hydrogens, 2 goes in front of the H2O, which yields:
- 1 CH|4| + O|2| \rightarrow CO|2| + 2 H|2|O
Inspection of the last atom to be balanced (oxygen) shows that the right hand side has four atoms, while the left hand side has two. It can be balanced by putting a 2 before O2, giving the balanced equation:
- CH|4| + 2 O|2| \rightarrow CO|2| + 2 H|2|O
This equation does not have any coefficients in front of CH4 and CO2, since a coefficient of 1 is dropped.
An ionic equation is a chemical equation in which electrolytes are written as dissociated ions. Ionic equations are used for single and double displacement reactions that occur in aqueoussolutions. For example in the following precipitation reaction:
- CaCl2(aq) + 2AgNO3(aq) \rightarrow Ca(NO3)2(aq) + 2AgCl(s)
the full ionic equation would be:
- Ca2+ + 2Cl− + 2Ag+ + 2NO3− \rightarrow Ca2+ + 2NO3− + 2AgCl(s)
and the net ionic equation would be:
- 2Cl−(aq) + 2Ag+(aq) \rightarrow 2AgCl(s)
or, in reduced balanced form,
- Ag+ + Cl− \rightarrow AgCl(s)
In this aqueous reaction the Ca2+ an
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Answers:steps to balance a redox reaction... 1) identify oxidation states 2) determine which is oxidized and which is reduced 3) write half reactions include electrons 4) balance electrons in half reactions 5) combine half reactions and cancel electrons 6) add counter ions 7) add THEN balance remaining species. ******** see if you can follow along... Ag is (0) Fe+2 is (+2) Ag+ is (+1) Fe is (0) Ag goes from 0 to +1 and is oxidized Fe goes from +2 to 0 and is reduced (reduction is reduction in charge fyi) Ag(0) ---> Ag(+1) + 1 e- Fe(+2) + 2e- ---> Fe(0) 2 Ag(0) ---> 2 Ag(+1) + 2 e- Fe(+2) + 2e- ---> Fe(0) 2 Ag(0) + Fe(+2) + 2e---> 2 Ag(+1) + 2 e- + Fe(0) 2 Ag(0) + Fe(+2) ---> 2 Ag(+1) + Fe(0) ***** and we can stop there... 2 Ag + Fe2+ 2 Ag+ + Fe
Answers:The formulas do get more complex in oxidation-reduction reactions, and there are a number of algebraic steps to take to balance the reaction. 1. Write the net ionic reaction for the oxidation step, then separately, for the reduction step. 2. Do atom balance, using H2O where necessary 3. Do electron balance, showing whether electrons are produced or used, after finding the change in oxidation state of the reacting species. 4. Do charge balance, adding H+ or OH- where necessary 5. Multiply whole ionic reaction(s) by number(s) which will cancel out the electrons. 6. Add up all the reactants and products, and cancel out those that show up on both sides of the equation.
Answers:You wrote: "KMnO4 + H2SO4 + NaHSO3 -> MnSO4" The reaction is between permanganate and bisulfite ions. Manganese is reduced to Mn2+. Add water and H+ to balance oxygen and hydrogen. Sulfur is oxidized from +4 to +6 in sulfate ion. Add water and H+ to balance oxygen and hydrogen. Make sure each half-reaction is charge balanced by including the electrons needed for oxidation or reduction. 2(8H+ + MnO4- + 5e- --> Mn2+ + 4H2O) 5(H2O + HSO3- --> SO4^2- + 3H+ + 2e-) ---------------------------------------------------------- 16H+ + 2MnO4- + 5HSO3- + 5H2O --> 2Mn2+ + 5SO4^2- + 8H2O + 15H+ simplify H+ + 2MnO4- + 5HSO3- --> 2Mn2+ + 5SO4^2- + 3H2O\ =========== Follow up ============== This is a redox reaction. There must be both oxidation and reduction. Reduction is the gain of electrons. This occurs with the manganese in permanganate ion. Oxidation is the loss of electrons. OILRIG Oxidation occurs to the sulfur in bisulfite ion as it goes from an oxidation number of +4 to +6. There must be an accompanying loss of two electrons, and those are the two electron which show up on the product side of the half-reaction.
Answers:Yeah, but it's HOW you get to the balanced equation that is important. The "best" way to balance redox reactions is by the "half-reaction" method. Write the balanced oxidation half-reaction and the balanced reduction half-reaction, multiply each by whatever coefficients are needed to make the electrons gained and lost the same and then add the two half-reactions together, and simply H+ and H2O if necessary. Since it is acid solution we use H+ and H2O to balance oxygen and hydrogen. 2 ( 8H+ + MnO4- + 5e- --> Mn2+ + 4 H2O 5 ( H2C2O4 --> 2CO2 + 2 H+ + 2e- ---------------------------------------------------------- 16H+ + 2MnO4- + 5H2C2O4 --> 2Mn2+ + 10CO2 + 8H2O + 10H+ simplify the H+ -- subtract 10 H+ from each side. 6H+ + 2 MnO4- + 5 H2C2O4 --> 2 Mn2+ + 10 CO2 + 8 H2O