Question

DO NOT ARBITRARILY BALANCE! In the following reactions you need to give both b anced half reactions in the appropriate acidic or basic solution: (1) one for oxidation and 2) o for reduction and then the 3) overall balanced equation. Do not break apart any of the follow ing ions or molecules. Balance in acidic solution: 1. MnOi(aq)+ H2C20,(aq) Mn" (aq) + CO2(g) 2. As S(s)+CIO5(aq) -H2ASO4 (aq)+SO (aq)+CT (aq) 3. Cr(CN) (aq)+ Ce(aq) Cr(aq)+ Ce (aq) + NO5(aq)+ CO2(g) 4. P,S,(aq)+ NO5(aq) H,PO,(aq) + SO (aq) + NOg) 5. As O(s)+ Mng(aq)AsO (aq)+Mn (aq) Balance in basic solution: 6. Cro (aq)+CIO (aq)Cro (ag) + Cl(g) Cr(aq) + Cro (aq)+ 10 (aq) 7. Cl(aq)+Crl(s) HCO5(aq) + VO(OH)2 (aq) 8. C,H,O,(ag) + V,O(ag) 9. Ce(aq) + Fe(SCN) (aq) Ce(OH)s (s) + Fe(OH)s (s) + SO (ag) + CO (aq)+ NO5(aq) 10. Al(s)+NO5(aq) AIO5(aq) + NH,(g)

Answer 1

Ans :-

Following steps while balancing the redox reaction by ion-exchange method :-

Step.1 :- Balance the atoms other than hydrogen and oxygen.

Step.2 :- Balance oxygen atoms by adding H₂O molecules in that side where oxygen content are less.

Step.3 :- Balance hydrogen atoms by adding H⁺ ion in that side where hydrogen content are less (For acidic).

Step.4 :- Balance charge by adding electrons in that side where positive charge is more.

Step.5 :- For basic equation add many number of OH^- as many H⁺ ions already present in acidic equation on both the sides of equation.

1. Given unbalanced redox reaction is :

MnO₄^- (aq) + H₂C₂O₄ (aq) --------------> Mn²⁺ (aq) + CO₂ (g)

(1). Oxidation-Half cell reaction is :

H₂C₂O₄ (aq) --------------> 2 CO₂ (g) + 2 H⁺ (aq) + 2e^-

(2). Reduction-Half cell reaction is :

MnO₄^- (aq) + 8 H⁺ (aq) + 5e^-  --------------> Mn²⁺ (aq) + 4 H₂O (l)

Multiply oxidation half cell reaction by 5 and reduction half cell reaction by 2, we have  

(3). Overall balanced redox reaction in acidic medium is :

2 MnO4- (aq) + 5 H2C2O4 (aq) + 6 H+ (aq) + --------------> 2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l)