Question

Balance the following REDOX reactions: 16. CrOz2 .....Cr+3 +Pb+2, (acid solution) + Pb

please solve this and show work.

Answer 1

Q16. Oxidation number of Cr in Cr₂O₇^2- is +6

Oxidation number of Cr in Cr³⁺ is +3

Since oxidation number of Cr decreases, therefore, Cr is reduced each Cr atom gains 3 electrons.

The balanced reduction half reaction is : Cr₂O₇^2- + 14 H⁺ + 6 e^-$\rightarrow$ 2 Cr³⁺ + 7 H₂O

Oxidation number of Pb in Pb (s) is 0

Oxidation number of Pb in Pb²⁺ is +2

Since oxidation number of Pb increases, therefore, Pb is oxidized and each Pb atom loses 2 electrons.

The balanced oxidation half reaction is : Pb $\rightarrow$ Pb²⁺ + 2 e^-

To eliminate the electron term e^-, the balanced oxidation half reaction is multiplied by a factor of 3

The balanced oxidation half reaction is : 3 Pb $\rightarrow$ 3 Pb²⁺ + 6 e^-

Now combine the balanced reduction half reaction and balanced oxidation half reaction to get the overall balanced equation

Cr₂O₇^2- + 14 H⁺ + 3 Pb $\rightarrow$ 2 Cr³⁺ + 7 H₂O + 3 Pb²⁺