chemistry question redox reaction

Hi, please could i have some help on this question?
I tried to balance the equation 8H+ + 3Sn2+ + MnO4 --> 3Sn4+ + Mn2+ +4 H20 and I think this is right but the ms says 5Sn2+ + 2MnO4 + 16H+ → 5Sn4+ + 2Mn2+ + 8H2O?
Thanks!

Reply 1

Nitrotoluene

Original post by anonymous56754

Do you use the ion-electron method, also known as the half-reaction method, to balance redox reactions? Please confirm.

(edited 4 months ago)

Reply 2

Mstew!

Sn(2+)—> Sn(4+) + 2e(-)
8H(+) + 5e(-) + MnO4(-)——> Mn(2+) + 4H20
These are the individual half equations: to combine you need to balance the electrons so they cancel out:
5(Sn(2+)—> Sn(4+) + 2e(-)) becomes
5Sn(2+)—> 5Sn(4+) + 10e(-)

2(8H(+) + 5e(-) + MnO4(-)——> Mn(2+) + 4H20) becomes
16H(+) + 10e(-) + 2MnO4(-)——> 2Mn(2+) + 8H2O

Now when you combine them the 10e(-) on either side cancel out:
5Sn(2+) + 16H(+) + 2MnO4(-) ——> 5Sn(4+) + 2Mn(2+) + 8H2O

Reply 3

TypicalNerd

Original post by anonymous56754

A good way of checking is looking at the charges on the ions on either side and seeing if they add up.

In the answer you got, the sum of the charges on the LHS should be (8 x +1) + (3 x +2) + (1 x -1) = +13. On the RHS, it is (3 x +4) + (1 x +2) = +14. These are not equal - you forgot to put the negative charge on MnO4^- and so this gave you a seemingly plausible answer.

I personally find the oxidation state method to be a good option for balancing equations like this and encourage people who struggle with half equations to use it. As MnO4^- features Mn in the +7 oxidation state and it gets reduced to the +2 oxidation state (Mn^2+), the coefficient of the reducing agent must be (+7) - (+2) = 5. As Sn^2+ is oxidised from the +2 oxidation state to +4 oxidation state, the coefficient of the reducing agent must be (+4) - (+2) = 2, so we have

2MnO4^- + 5Sn^2+ + …H^+ —> …Mn^2+ + …Sn^4+ + …H2O

This should be quite straightforward to balance now. See how there are 8 oxygens on the LHS, so there must be 8 waters produced, implying you need 16H^+? Similarly, since you use up 2 MnO4^- ions and 5 Sn^2+ ions, you must make 2 Mn^2+ ions and 5 Sn^4+ ions. Thus, you get

2MnO4^- + 5Sn^2+ + 16H^+ —> 2Mn^2+ + 5Sn^4+ + 8H2O

As required.

(edited 4 months ago)

Reply 4

anonymous56754

Thank you everyone for your replies, I realised I missed out the - sign on MnO4- but I’ve reached the correct answer now!