## Related questions with answers

Balance each of the following oxidation-reduction reactions. For each, indicate which substance is being oxidized and which is being reduced.

a. $\mathrm{Co}(s)+\mathrm{Br}_2(l) \rightarrow \mathrm{CoBr}_3(s)$

Solution

VerifiedTo solve this problem we use rules for balancing equations that say: first start with the most complicated molecule and then proceed element by element and use appropriate coefficient to have the same number of the element on both sides of the arrow.

Coefficients used are the smallest integers that balance the number of each element on both side of the arrow.

Unbalanced equation is:

$\mathrm{Co_ { ( s ) } + Br _ { 2 ( l ) } \rightarrow CoBr _ {3 ( s )}}$

The most complicated compound is $\mathrm{CoBr_3}$.

The easiest way to solve this problem is to use coefficient 2 before $\mathrm{CoBr_3}$.

If we use coefficient 2 before $\mathrm{CoBr_3}$, the number of Co atoms on the right side of the reaction arrow is 2.

To balance number of Co atoms on both sides of the reaction arrow we must use coefficient 2 before $\mathrm{Co}$

$\mathrm{2Co_ { ( s ) } + Br _ { 2 ( l ) } \rightarrow 2CoBr _ {3 ( s )}}$

The number of $\mathrm{Br}$ atoms on the right side of the reaction arrow is 6.

To balance a number of $\mathrm{Br}$ atoms on both sides of the reaction arrow we must use coefficient 3 before $\mathrm{Br_2}$.

Balanced equation:

$\mathrm{2Co_ { ( s ) } + 3Br _ { 2 ( l ) } \rightarrow 2CoBr _ {3 ( s )}}$

## Create an account to view solutions

## Create an account to view solutions

## More related questions

1/4

1/7