In + Cr2O72− → In3+ + Cr3+ Reduction, oxidation, and net

Answer 1

Half reactions:

In ==> In³⁺ + 3e- ... oxidation half reaction (balanced)

Cr₂O₇^2- ==> Cr³⁺ ... reduction half reaction (unbalanced)

Cr₂O₇^2- ==> 2Cr³⁺ ... balanced for Cr

Cr₂O₇^2- ==> 2Cr³⁺ + 7H₂O ... balanced for O

Cr₂O₇^2- + 14H⁺ ==> 2Cr³⁺ + 7H₂O ... balanced for H

Cr₂O₇^2- + 14H⁺ + 6e- ==> 2Cr³⁺ + 7H₂O ... final balanced reduction equation

In ==> In³⁺ + 3e- (x2 in order to equalize electrons in oxidation and reduction 1/2 reactions)

2In ==> 2In³⁺ + 6e- ... final balanced oxidation reaction

Add the oxidation and reduction reactions together to get the final balanced redox equation:

Cr₂O₇^2- + 14H⁺ + 2 In ===> 2Cr³⁺ + 7H₂O + 2 In³⁺ ... FINAL NET BALANCED EQUATION