Write a balanced equation for the following reduction-oxidation reaction.
SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
can someone help me step by step??
Answered above.
SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
first deal with S, Mn atoms
there must be the same number on each side
nSO3(2-) + mMnO4(-) --- nSO4(2–) + mMn(2+)
now try 2jH(+) on the left and jH2O on the right
nSO3(2-) + mMnO4(-) +2jH(+) --- nSO4(2–) + mMn(2+) + jH2O
O atoms
3 n + 4 m = 4 n + j
or
n+j = 4 m
+ charges
-2 n - 1 m + 2 j = -2 n + 2 m
so from the second one
3 m = 2 j
j = (3/2) m
from the first one
n + (3/2) m = 4 m
or
n = (5/2) m
Now go back using m = 2 to make the fraction go away
then j = 3
and n = 5
5 SO3(-2) + 2 MnO4(-1) + 6 H(+1) --> 3 H2O + 5 SO4(-2) + 2 Mn(+2)
Sure, I can help you step by step.
To write a balanced equation for a reduction-oxidation (redox) reaction, you need to follow these steps:
1. Identify the oxidation numbers of each element in the reaction. In this case, let's assign oxidation numbers to the elements involved:
SO3(2-) --> The oxidation number of S is +6, and the oxidation number of O is -2. Since there are three oxygen atoms, the total oxidation number for O is -6. So, to balance the charges, the oxidation number for S should be +4.
MnO4(-) --> The oxidation number of O is -2, and there are four oxygen atoms. So, the total oxidation number for O is -8. The overall charge of MnO4(-) is -1. To find the oxidation number for Mn, we can use the formula: charge = sum of oxidation numbers. So, (-1) = Mn + (-8). Solving for Mn, we get the oxidation number of Mn as +7.
SO4(2-) --> The oxidation number of O is -2, and there are four oxygen atoms. So, the total oxidation number for O is -8. The overall charge of SO4(2-) is -2. To find the oxidation number for S, we can use the formula: charge = sum of oxidation numbers. So, (-2) = S + (-8). Solving for S, we get the oxidation number of S as +6.
Mn(2+) --> The oxidation number of Mn is +2.
2. Separate the reaction into two half-reactions, one for the oxidation reaction and another for the reduction reaction.
Oxidation half-reaction: SO3(2-) --> SO4(2–) + 2e-
Reduction half-reaction: MnO4(-) + 8H+ + 5e- --> Mn(2+) + 4H2O
3. Balance the number of atoms in each half-reaction (excluding hydrogen and oxygen).
Oxidation half-reaction: SO3(2-) --> SO4(2–) + 2e-
Reduction half-reaction: MnO4(-) + 8H+ + 5e- --> Mn(2+) + 4H2O
4. Balance the hydrogen atoms by adding H+ ions to the appropriate side of each half-reaction.
Oxidation half-reaction: SO3(2-) --> SO4(2–) + 2e-
Reduction half-reaction: MnO4(-) + 8H+ + 5e- --> Mn(2+) + 4H2O
5. Balance the charge on each half-reaction by adding electrons (e-) to the appropriate side.
Oxidation half-reaction: SO3(2-) --> SO4(2–) + 2e-
Reduction half-reaction: MnO4(-) + 8H+ + 5e- --> Mn(2+) + 4H2O
6. Equalize the number of electrons in both half-reactions.
Oxidation half-reaction: 5SO3(2-) --> 5SO4(2–) + 10e-
Reduction half-reaction: 2MnO4(-) + 16H+ + 10e- --> 2Mn(2+) + 8H2O
7. Multiply each half-reaction by an appropriate coefficient to equalize the number of electrons.
Oxidation half-reaction: 5SO3(2-) --> 5SO4(2–) + 10e-
Reduction half-reaction: 2MnO4(-) + 16H+ + 10e- --> 2Mn(2+) + 8H2O
8. Add the two balanced half-reactions together.
5SO3(2-) + 2MnO4(-) + 16H+ --> 5SO4(2–) + 2Mn(2+) + 8H2O
This is the balanced equation for the given redox reaction:
5SO3(2-) + 2MnO4(-) + 16H+ --> 5SO4(2–) + 2Mn(2+) + 8H2O