balance the ionic equation: Al + H+------------------> A^3 + H2
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The equation is balanced asAl → Al3+ + 3e− −−−−−12H+ +2 e− → H2 −−−−−−−2Balance the number of electrons on both the equtions.I equation is multiplied by 2 and II equation by 3 and add the two new equation2Al →2 Al3+ + 6e−6H+ +6 e− → 3H2−−−−−−−−−−−2 Al + 6 H+ →2Al3+ + 3 H2−−−−−−−−−−−−−
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considering the reactions into half reactions...which is also known as redox recation
Al --> Al^+3 loses 3 electrons
2 H^+1 --> H2 gains 2 electrons
The electrons must be balanced
While you could write Al + 3H^+1 --> Al^+3 + 1.5 H2 (3 electrons lost by Al --> Al^+3, 3 electrons gained by 3 H^+1 going to 1.5 H2)
However, equations must be written with whole numbers; you can't have 1.5 molecules of H2, so you must - in this case - double all the numbers to get
2 Al + 6 H^+1 --> 2 Al^+3 + 3 H2
Al --> Al^+3 loses 3 electrons
2 H^+1 --> H2 gains 2 electrons
The electrons must be balanced
While you could write Al + 3H^+1 --> Al^+3 + 1.5 H2 (3 electrons lost by Al --> Al^+3, 3 electrons gained by 3 H^+1 going to 1.5 H2)
However, equations must be written with whole numbers; you can't have 1.5 molecules of H2, so you must - in this case - double all the numbers to get
2 Al + 6 H^+1 --> 2 Al^+3 + 3 H2
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