Question

Assume a reaction takes place in a basic solution to form the given products:MnO4–(aq) + Cl–(aq) --> MnO2(s) + Cl2(g) (unbalanced). Do each step individually:1. Balance the given half-reactions for atoms and charge.MnO4– + H2O -->MnO2 + OH– andCl– --> Cl2.2. Multiply to balance the charges in the reaction.3. Add the equations and simplify to get a balanced equation.

Answer 1

Answer:

$6Cl^-(aq)+2MnO_4^-(aq)+4H_2O(l)\operatorname{\rightarrow}3Cl_2(g)+2MnO_2(s)+8OH^-(aq)$

Explanations:

What is redox reaction?

Redox reaction is the reaction that occurs between an oxidizing and reducing substrates.

Given the chemical equation below;

$MnO_4^-(aq)+Cl^-\rightarrow MnO_2(s)+Cl_2(g)$

Separate into half reactions

$\begin{gathered} MnO_4^-\rightarrow MnO_2 \\ Cl^-\rightarrow Cl_2 \end{gathered}$

1) Balance the charges by adding the equivalent electrons

$\begin{gathered} MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^- \\ 2Cl^-\rightarrow Cl_2+2e^- \end{gathered}$

2) Multiply to balance the charge

$\begin{gathered} MnO_4^-+2H_2O+3e^-\operatorname{\rightarrow}MnO_2+4OH^-\text{ }\times2 \\ 2Cl^-\rightarrow Cl_2+2e^-\text{ }\times3 \\ _{-------------------------------------} \\ 2MnO_4^-+4H_2O+6e^-\operatorname{\rightarrow}2MnO_2+8OH^- \\ 6Cl^-\rightarrow3Cl_2+6e^- \end{gathered}$

3) Add the equations and simplify to get a balanced equation to have;

$6Cl^-(aq)+2MnO_4^-(aq)+4H_2O(l)\rightarrow3Cl_2(g)+2MnO_2(s)+8OH^-(aq)$

This gives the balanced equation.