Q4.95P - Chemistry: Molecular Nature Of Matter And Change

Question

Predict the product(s) and write a balanced equation for each of the following redox reactions:

(a) Cesium + Iodine →

(b) Aluminum + aqueous Manganese(II) sulfate →

(c) Sulfur dioxide + Oxygen →

(d) Butane and Oxygen →

(e) Write a balanced net ionic equation for (b).

 

Step-by-Step Solution

Answer

The product and the balance equation of the given reactions are:

(a)$2Cs\left(s\right)+2l_2\left(s\right)\to 2CsI\left(s\right)$

(b)$2Al\left(s\right)+3MnS{O}_4\left(aq\right)\to A{l}_2{\left(S{O}_4\right)}_3\left(aq\right)+3Mn\left(s\right)$

(c)$2S{O}_2\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2S{O}_3\left(g\right)$

(d)$2C_4{H}_{10}\left(g\right)+13O_2\left(g\right)\stackrel{}{\to }8C{O}_2\left(g\right)+10H_{2}O\left(l\right)$

(e)$2Al\left(s\right)+3M{n}^{2+}\left(aq\right)\to 2A{l}^{3+}\left(aq\right)+3Mn\left(s\right)$

 

Step 1: Determine the product of the equation (a)

The given redox reaction is a combination reaction.

Cesium combines with iodine and forms cesium iodide.

 

The balanced reaction is:

$2Cs\left(s\right)+2l_2\left(s\right)\to 2CsI\left(s\right)$

Step 2: Determine the product of the equation (b)

The balanced redox reaction is a displacement reaction.

 $2Al\left(s\right)+3MnS{O}_4\left(aq\right)\to A{l}_2{\left(S{O}_4\right)}_3\left(aq\right)+3Mn\left(s\right)$

Step 3: Determine the product of the equation (c)

The given redox reaction is a combination reaction.

Sulfur dioxide combines with oxygen and forms sulfur trioxide.

 

The balanced reaction is:

$2S{O}_2\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2S{O}_3\left(g\right)$

Step 4: Determine the product of the equation (d)

The given redox reaction is a combustion reaction.

Combustion of butane takes place in presence of oxygen and forms carbon dioxide and water.

 

 

 

The balanced reaction is:

$2C_4{H}_{10}\left(g\right)+13O_2\left(g\right)\stackrel{}{\to }8C{O}_2\left(g\right)+10H_{2}O\left(l\right)$

Step 5: Determine the total ionic equation

The balanced equation is:

$2Al\left(s\right)+3MnS{O}_4\left(aq\right)\to A{l}_2{\left(S{O}_4\right)}_3\left(aq\right)+3Mn\left(s\right)$

Now, the total ionic equation is

$2Al\left(s\right)+3M{n}^{2+}\left(aq\right)+3S{O}_4^{2-}\left(aq\right)\to 2A{l}^{3+}\left(aq\right)+3S{O}_4^{2-}\left(aq\right)+3Mn\left(s\right)$

Step 10: Determine the net ionic equation

$S{O}_4^{2-}$is the spectator ion in the total ionic equation. As you know that spectator ion is not present in the net ionic equation. Hence the net ionic equation is:

$2Al\left(s\right)+3M{n}^{2+}\left(aq\right)\to 2A{l}^{3+}\left(aq\right)+3Mn\left(s\right)$