Question

Consider this mechanism:
Step 1: ClO-(aq) + H2O(l) HClO(aq) + OH-(aq) (fast)
Step 2: I-(aq) + HClO(aq) HIO(aq) + Cl-(aq) (slow)
Step 3: OH-(aq) + HIO(aq) H2O(l) + IO-(aq) (fast)

What is the overall reaction?
A.
HClO (aq) + H2O (l) OH- (aq) + Cl- (aq)
B.
I- (aq) + HClO (aq) HIO (aq) + Cl- (aq)
C.
OH- (aq) +HIO (aq) H2O (l) + IO- (aq)
D.
ClO- (aq) + I- (aq) Cl- (aq) + IO- (aq)
E.
ClO- (aq) + H2O(l) HClO(aq) + OH- (aq)

Answer 1

Answer:

$ClO^- + I^-\rightarrow IO^- + Cl^-$

Explanation:

Given a reaction mechanism, we will typically have catalysts and intermediates which will not be observed in the final balanced overall reaction. In order to obtain a net reaction, we need to add all the separate steps of the given reaction.

First of all, the reactants are combined followed by a combination of products. The repeating species on both sides of the final equation are then canceled out. Let's sum everything we have in our steps:

$ClO^- + H_2O + I^- + HClO + OH^- + HIO\rightarrow HClO + OH^- + HIO + Cl^- + H_2O + IO^-$

Repeating species that can be canceled out are:

$H_2O,~HClO,~OH^-,~HIO$

This leaves a net reaction of:

$ClO^- + I^-\rightarrow IO^- + Cl^-$