Im totally lost on how to do redox reactions. Can any one show me the steps on how to do MnO4-1 +C7H60 => Mn2 +c7H6O2?
im trying to study for a test and im totally lost.
thanks in advanced.
MnO4- +5e -> Mn2+}*2
C+1 -2e -> C+3 }*5
2MnO4- + 5C6H7O +6H+ -> 2Mn2+ + 5C6H7O2 + 3H2O
Step 1: realize the oxidation of element: (oxidation of H and O not change)
write half reaction
equalize the number of election yieled and accepted by the coefficient ( here e yield is 2 and e accepted is 5)
let the coefficient into the equation
check the equation: if the side of equation has more O: add H+ and the H2O to the other side
the last step: check the charge of both side, if wrong check every step you did again.
for your equation, becaue the carbons on the ring do not change their oxidation state, only carbon of carbonyl group change because aldehyde group is converted into carboxyl group
I think you have to review the following conceps below :
In free elements (that is, in uncombined state), each atom has an oxidation number of zero. Ex. In O2, the oxidation number of each oxygen atom is zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Ex. The oxidation number of Ca2+ is +2.
3. All alkali metals (elements in column 1of the periodic table, with the exception of hydrogen) have an oxidation number of +1. Ex. The oxidation numbers of Li, K, and Na will always be +1.
4. All alkaline earth metals (elements in column 2 of the periodic table) have an oxidation number of +2. Ex. The oxidation number of Ba is +2.
5. The oxidation number of Aluminum (Al) is always +3.
6. The oxidation number of oxygen in most compounds (such as H2O and CO2) is -2. In hydrogen peroxide (H2O2) and peroxide (O22-) oxygen shows a -1 oxidation number.
7. The oxidation number of hydrogen is +1, except when in is bonded to a metal as a negative ion, in which case it is -1. Ex. H2O shows hydrogen as +1. NaH shows hydrogen as -1.
8. When halogens (elements in column 17 on the periodic table) form negative ions, they will have an oxidation number of -1. Ex. NaCl and CaCl2 both show chlorine with a -1 oxidation number.
9. In a neutral molecule, the sum of the oxidation numbers of all of the atoms must be zero. Ex. In H2O, each hydrogen is +1 and the oxygen is -2. So, (2 x +1) + (-2) = 0.
10. In a polyatomic ion, the sum of oxidation numbers of all the elements in the ion must be equal to the net charge of the ion. Ex. In the polyatomic ion known as hydroxide (OH-), the oxygen is -2 and the hydrogen is +1. So, (-2) + (+1) = -1, the same as the charge on the hydroxide ion (OH-)
good luck
Sure! Redox reactions involve the transfer of electrons between species. In order to balance a redox reaction, you can follow a simple step-by-step process.
Step 1: Determine the oxidation states of each atom in the reaction. In this case, we have MnO4-1 (permanganate ion) and C7H6O (benzoic acid).
The oxidation state of Mn in MnO4- is +7, and the oxidation state of O is -2. The total charge of MnO4- is -1, so the oxidation state of oxygen in this case is (-2) x 4 = -8. Therefore, the oxidation state of Mn in MnO4- is +7 because the sum of the oxidation states in a neutral molecule must be zero.
The oxidation state of C in C7H6O2 is +3, and the oxidation state of O is -2. The total charge of C7H6O2 is zero, so the sum of the oxidation states in this compound is equal to zero.
Step 2: Identify which atoms are being reduced and which atoms are being oxidized. In a redox reaction, oxidation is the loss of electrons, and reduction is the gain of electrons.
In this case, Mn is being reduced from +7 to +2, so it is gaining electrons. C is being oxidized from +3 to +4, so it is losing electrons.
Step 3: Write half-reactions for the oxidation and reduction separately. In the reduction half-reaction, write the species being reduced and the number of electrons gained. In the oxidation half-reaction, write the species being oxidized and the number of electrons lost.
Reduction half-reaction: MnO4- + 5e- => Mn2+
Oxidation half-reaction: C7H6O2 => C7H4O2 + 2e-
Step 4: Balance the number of electrons in both half-reactions. In order to do this, multiply the oxidation half-reaction by a suitable number so that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction.
Multiply the oxidation half-reaction by 5:
5 x (C7H6O2 => C7H4O2 + 2e-) gives you: 5C7H6O2 => 5C7H4O2 + 10e-
Now the electrons lost in the oxidation half-reaction (10e-) are equal to the electrons gained in the reduction half-reaction (5e-).
Step 5: Balance the other atoms in each half-reaction. In the reduction half-reaction, balance the Mn atoms and O atoms. In the oxidation half-reaction, balance the C and H atoms.
Reduction half-reaction: MnO4- + 5e- => Mn2+
The Mn atoms are already balanced, but there are 4 O atoms on the left side and 2 O atoms on the right side. Add 2H2O to the left side to balance the O atoms:
MnO4- + 5e- + 8H+ => Mn2+ + 4H2O
Oxidation half-reaction: 5C7H6O2 => 5C7H4O2 + 10e-
The C and H atoms are already balanced.
Step 6: Balance the overall charge by adding or subtracting electrons from one of the half-reactions. In this case, the reduction half-reaction has a net negative charge, so we need to add electrons to balance the charge.
Multiply the reduction half-reaction by 2 to balance the overall charge:
2MnO4- + 10e- + 16H+ => 2Mn2+ + 8H2O
Step 7: Add the two half-reactions together. Make sure that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction.
2MnO4- + 10e- + 16H+ + 5C7H6O2 => 2Mn2+ + 8H2O + 5C7H4O2
And there you have it! The balanced redox reaction is:
2MnO4- + 16H+ + 5C7H6O2 => 2Mn2+ + 8H2O + 5C7H4O2