Question

Balance each of the following in BASE: Show work

Answer 1

Answer : The balanced chemical equation in a basic solution are,

(A) $2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) $2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) $4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) $Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

Explanation :

(A) The given chemical reaction is :

$CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)$

The oxidation-reduction half reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}$

or,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) The given chemical reaction is :

$NO_2(g)\rightarrow NO_3^-(aq)+NO_2^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $NO_2+2OH^-\rightarrow NO_3^-+H_2O+1e^-$

Reduction : $NO_2+1e^-\rightarrow NO_2^-$

The electrons in both the reactions are balanced. Now added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) The given chemical reaction is :

$Zn(s)+NO_3^-(aq)\rightarrow Zn(OH)_4^{2-}(aq)+NH_3(g)$

The oxidation-reduction half reaction will be :

Oxidation : $Zn+4OH^-\rightarrow [Zn(OH)_4]^{2-}+2e^-$

Reduction : $NO_3^-+6H_2O+8e^-\rightarrow NH_3+9OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 4 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) The given chemical reaction is :

$Br_2(l)\rightarrow BrO_3^-(aq)+Br^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Br_2+12OH^-\rightarrow 2BrO_3^-+6H_2O+10e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^-$

In order to balance the electrons, we multiply the reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$