Question

Balance this reaction that occurs in an acidic solution: TcO4 − + Sn 2+ → Tc 4+ + Sn4+

May you please explain each step? Thank you

Answer 1

First we will write the two half reactions-

TcO4^- ---> Tc⁴⁺

Sn²⁺ ----> Sn⁴⁺

O.N. of Tc in TcO4^- = x + 4*(-2) = -1 = x = +7

First we will balance the charges on both sides-

Tc goes from +7 to +4, so we will add 3e^- on the left side so that both side has +4 charge-.

TcO4^- + 3e- ----> Tc⁴⁺

Similarly we will add 2 e^- on the right side for the Sn half reaction-

Sn²⁺ ---> Sn⁴⁺ + 2e^-

Now we will balance H and O-

In the Tc half reaction there are 4 O on the left side so to balance that we will add 4H2O on the right side -

TcO4^- + 3e^- ---> Tc⁴⁺ + 4H2O

Now to balance the H, we will add 8H⁺ on the left side so we get-

TcO4^- + 3e^- + 8H⁺ ---> Tc⁴⁺ + 4H2O

Sn half reaction does not have any H or O so no need to do these steps-

Sn²⁺ ----> Sn⁴⁺ + 2e^-

Now we will multiply Tc half reaction by 2 and Sn half reaction by 3, so that both reactions have 6 electrons which get cancelled when they are added up, so we get-

2TcO4^- + 6e^- + 16H⁺---> 2Tc⁴⁺ + 8H2O

3Sn²⁺ ---> 3Sn⁴⁺ + 6e^-

Adding them both we get-

2TcO4^- + 3Sn²⁺ + 16H⁺---> 2Tc⁴⁺ + 3Sn⁴⁺ + 8H2O