If the reaction can be balanced, you will find coefficients. Probably the most important characteristics of the algebraic method is that - contrary to the inspection method - it is guaranteed to give you an answer. Let's put a = 1 and calculate all other coefficients simply using already known values:Ĭr 2O 7 2- + 14H + + 6Fe 2+ → 2Cr 3+ + 7H 2O + 6Fe 3+ Let's get rid of b and d in the charge balance equation: (note that sign of coefficients in the last equation depends on the charge sign). However, charge has to be balanced as well, and that will give us last equation needed to balance reaction: For algebraic method we can have one equation less than variables - so we are still one equation short. What equations do we have? Four balances of atoms:īut that's not enough to balance equation - we have six coefficients and four equations. Let's try it forĪCr 2O 7 2- + bH + + cFe 2+ → dCr 3+ + eH 2O + fFe 3+ It's also easy to use the algebraic method to balance redox reaction with charged species. Imagine finding them by inspection method! In this case the smallest common denominator is 10, so if we multiply all numbers by 10 we get:Īnd you may check that these are the correct coefficients. These are hardly integer, but all we have to do is to find the smallest common denominator to have a list of integer coefficients in numerators. Assuming a = 1 and simply substituting calculated values we have: c = 4×e and d = 4×a are substitutions that we are about to use to reduce number of unknowns:įor someone fluent in algebra it is obvious that we have already finished - it is now enough to assume that one of the variables equals 1 to calculate values of all others. We need five coefficients, and there are only four equations (one for each element present) - but it shouldn't bother us, as we know that we have additional information that works as an additional equation.ĪP 2I 4 + bP 4 + cH 2O → dPH 4I + eH 3PO 4īalances for iodine and oxygen make this set look much easier than expected. Looks easy but soon gets surprisingly hard and the coefficients become pretty high, which makes you wonder if you have not made some mistake (see some balancing hints at the bottom of that page). This first example doesn't look convincing - why do we have to solve set of equations when the reaction equation can be easily balanced by other means? Good point - but what if the reaction can be not easily balanced? To find them we can assume one of the coefficients to be 1: In algebra it usually means that the set of equations doesn't have a unique solution, but in the case of chemical equations we have one additional information - all coefficients must be integer and they must be the smallest ones. In this case we have very simple equation a = 6×b that we can use to substitute 6×b for a in the second and third equation to get:īoth equations are identical. Such equation sets is not a thing that you may want to solve manually, although when balancing chemical equations in most cases it can be done relatively easy, as most equations don't contain all unknowns. Quite often you will end with many more equations and many more unknows. We have three equations, and three unknowns - nothing particularly difficult to solve. We can write similar equations for the remining elements - hydrogen:Īs there are no free terms in this set of equations, it has a trivial solution (a = b = c = 0) which we are not interested in. Using coefficients a, b and c we can tell that we have 1×a atoms of boron on the left (one atom per each H 3BO 3 molecule), and 6×b + 0×c on the right (6 atoms of boron per each H 4B 6O 11 molecule and no boron in water). Our reaction has three coefficients a, b and c: What does 'balanced' mean? It means that for every element, there is the same number of atoms on both sides of the reaction equation. It can be rather easily balanced by inspection, but let's try a more systematic approach. That's the method EBAS - our chemical reaction equation balancer - uses. General algebraic method of balancing chemical reaction equationsĪpart from the three already described methods, there is also a general method, often less user friendly - but thanks to its systematic approach perfect for use in computer programs.
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