Question

Write the balanced equations for the following reactions, using the half reaction method.

Please note that you must write down all the steps that you took to balance the reaction in detail and explain your working.

 

Cr(OH)₃ + Br₂ → CrO₄^2- + Br^- in basic solution

Answer 1

Step 1. Identify and write out all redox couples in reaction

$O:Cr(OH)_3 → CrO_4^{2-} + 3e^{-}$

$R:Br_2 + 2e^- → 2Br^-$

Step 2: Balance the charge. For reactions in a basic solution, balance the charge so that both sides have the same total charge by adding an OH- ion to the side deficient in negative charge.

$O: Cr(OH)_3+5OH^{-}\rightarrow CrO_4^{2-}+3e^{-}$

 $R: Br_2+2e^-\rightarrow 2Br^-$

Step 3:Balance the oxygen atoms by adding water molecules.

$O: Cr(OH)_3+5OH^{-}\rightarrow CrO_4^{2-}+3e^{-}+4H_2O~~~~~~-(1) \\ R: Br_2+2e^-\rightarrow 2Br^-~~~~~~~-(2)$

Step 4: Make electron gain equivalent to electron lost,To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

  Multiplying eqs.(1) by 2 and eqs.(2) by 3 and we get-

$O: 2Cr(OH)_3+10OH^{-}\rightarrow 2CrO_4^{2-}+6e^{-}+8H_2O\\ R: 3Br_2+6e^-\rightarrow 6Br^-$

Step 5:Add the half-reactions together. 

  $2Cr(OH)_3+10OH^{-}+3Br_2+6e^-\rightarrow 6Br^-+ 2CrO_4^{2-}+6e^{-}+8H_2O$

Step 6: Simplify the equation-

 

  $2Cr(OH)_3+10OH^{-}+3Br_2\rightarrow 6Br^-+ 2CrO_4^{2-}+8H_2O$