Balance the equation in aqueous acidic solution:

C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)

please teach me how to do it!

There are two ways to balance equations

1. by inspection
2. by maths

The first is usually quicker IF you can see how to do it. The second can take longer but always works.

So to balance by inspection:

C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + CO2(g)

First make sure you are working with the correct equation, which here is

H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + CO2(g) + H2O

1. leave atoms (or ions) that are on their own to last, so here leave the Mn2+ to last

2. include fractions if you want to but remember that the final equaion must not contain fractions

3. Another things to notice is that in the starting equation there is an even numebr of O atoms on the LHS and an odd number on the RHS, so start by sorting this out

H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + CO2(g)+ 2H2O

4. Then balancing the H atoms

4H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + CO2(g)+ 2H2O

5. and balance the C atoms

4H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + 2CO2(g)+ 2H2O

There are now 8 O atoms on the LHS and 6 on the RHS which we can correct by putting a 4 in front of the H2O and rebalancing the H atoms

8H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + 2CO2(g)+ 4H2O

Which balances in terms of atoms but the charges do not balance as there is 5+ on the LHS and 2+ on the RHS. To balance this we need 2.5 of the C2O42- as this then gives

(8+)+(5-)+(1-) = 2+ on the LHS which is the same as the RHS.

So putting in the 2.5 and balancing the C atoms on the RHS.

8H+ + 2.5C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + 5CO2(g)+ 4H2O

but we are not allowed the fraction so

16H+ + 5C2O42-(aq) + 2MnO4-(aq) �¨ 2Mn2+(aq) + 10CO2(g)+ 8H2O

OK so how do we do this by maths?

start by inserting symbols in front of the species, where the numbers will go

H+ + C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + CO2(g) + H2O

aH+ + bC2O42-(aq) + cMnO4-(aq) �¨ Mn2+(aq) + dCO2(g) + eH2O

Note I have left one (Mn2+) as 1, but it does not matter which you choose.

Next you can write equations to balance the atoms

H. 2e=a
C. 2b=d
O. 4b+4c=2d+e
Mn. c=1

and the charge
a-2b-c=2

It is then some trivial algebra to solve these

c=1
e=4
a=8
b=5/2
d=5

8H+ + 5/2C2O42-(aq) + MnO4-(aq) �¨ Mn2+(aq) + 5CO2(g) + 4H2O

then remove the fraction by multiplying both side by 2 we get the same equation as by inspection

16H+ + 5C2O42-(aq) + 2MnO4-(aq) �¨ 2Mn2+(aq) + 10CO2(g)+ 8H2O

SWAG

Schwanz ^^

writing method of this question is totally wrong

Sure! I'd be happy to explain how to balance this equation in aqueous acidic solution.

Step 1: Identify the unbalanced elements and oxidation states of each element in the equation.

In this equation, we have:
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)

The unbalanced elements are carbon (C) and oxygen (O).

Carbon (C) is being oxidized from an oxidation state of +3 in C2O42- to +4 in CO2.
Oxygen (O) is being reduced from an oxidation state of -2 in C2O42- to -2 in CO2.

Step 2: Balance the atoms in the half-reactions.

a) Half-Reaction for Reduction (MnO4- → Mn2+)
MnO4-(aq) → Mn2+(aq)

First, balance the manganese (Mn) atoms:
MnO4-(aq) → Mn2+(aq)

Next, balance the oxygen (O) atoms by adding water (H2O):
MnO4-(aq) + 4H2O(l) → Mn2+(aq)

Finally, balance the charge by adding electrons (e-):
MnO4-(aq) + 4H2O(l) + 8e- → Mn2+(aq)

b) Half-Reaction for Oxidation (C2O42- → CO2)
C2O42-(aq) → CO2(g)

First, balance the carbon (C) atoms:
C2O42-(aq) → 2CO2(g)

Next, balance the oxygen (O) atoms by adding water (H2O):
C2O42-(aq) + 2H2O(l) → 2CO2(g)

Finally, balance the charge by adding electrons (e-):
C2O42-(aq) + 2H2O(l) + 4e- → 2CO2(g)

Step 3: Multiply the half-reactions to balance the number of electrons transferred.

To balance the number of electrons transferred, we need to multiply the first half-reaction by 5 and the second half-reaction by 2:

5(MnO4-(aq) + 4H2O(l) + 8e-) → 5Mn2+(aq) + 20OH-(aq)
2(C2O42-(aq) + 2H2O(l) + 4e-) → 4CO2(g)

Now, the number of electrons in each half-reaction is balanced.

Step 4: Combine the half-reactions and cancel out common species.

To combine the two half-reactions, we need to make the number of electrons equal in both reactions. By multiplying the first half-reaction by 2 and the second half-reaction by 5, we get:

10(MnO4-(aq) + 4H2O(l) + 8e-) → 10Mn2+(aq) + 40OH-(aq)
5(C2O42-(aq) + 2H2O(l) + 4e-) → 10CO2(g)

Finally, we can cancel out the common species:

10MnO4-(aq) + 20H2O(l) + 16e- → 10Mn2+(aq) + 40OH-(aq)
5C2O42-(aq) + 10H2O(l) + 20e- → 10CO2(g)

Step 5: Balance the hydrogen (H) by adding H+ ions.

Since the solution is acidic, we need to balance hydrogen (H) atoms by adding H+ ions:

10MnO4-(aq) + 20H2O(l) + 16e- → 10Mn2+(aq) + 40OH-(aq)
5C2O42-(aq) + 10H2O(l) + 20e- → 10CO2(g) + 20H+(aq)

Step 6: Balance the charge by adding electrons (e-).

To balance the charges, we add electrons (e-) to the appropriate side of the equation. In this case, we need to add 16e- to the right side of the first half-reaction and 20e- to the left side of the second half-reaction:

10MnO4-(aq) + 20H2O(l) + 16e- → 10Mn2+(aq) + 40OH-(aq)
5C2O42-(aq) + 10H2O(l) + 20e- → 10CO2(g) + 20H+(aq) + 20e-

Step 7: Simplify and combine the half-reactions.

Now, we can simplify the equation by canceling out the electrons:

10MnO4-(aq) + 20H2O(l) + 16e- → 10Mn2+(aq) + 40OH-(aq)
5C2O42-(aq) + 10H2O(l) + 20e- → 10CO2(g) + 20H+(aq) + 20e-

The final balanced equation is:

10MnO4-(aq) + 5C2O42-(aq) + 44H+(aq) → 10Mn2+(aq) + 10CO2(g) + 22H2O(l) + 20OH-(aq)

And that's how you balance the equation in aqueous acidic solution.

This is not right. Do not use this answer on Mastering Chemistry.