2 moles of water are produced. Here, hydrogen is the limiting reagent. 1 mole of oxygen is used up with 2 moles of hydrogen to form 2 moles of water. 1 moles of oxygen remains.

4 0

4 years ago

10 points) Given the following two half-reactions, write the overall reaction in the direction in which it is product-favored, a

IceJOKER [234]

Answer:

The over all reaction :

$2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)$

The standard cell potential of the reaction is 0,.897 Volts.

Explanation:

Reduction at cathode :

$Fe^{3+}(aq)+e^-\rightarrow Fe^{2+}(s)$ ..[1]

$E^o_{Fe^{3+}/Fe^{2+}}=0.771 V$

Reduction potential of $Fe^{3+}$ to $Fe^{2+}=0.771 V$

Oxidation at anode:

$Pb(s)\rightarrow Pb^{2+}(aq)+2e^-$ .[2]

$E^o_{Pb^{2+}/Pb}=-0.126 V$

Reduction potential of $Pb^{3+}$ to $Pb=-0.126 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=E^o_{Fe^{3+}/Fe^{2+}}-E^o_{Pb^{2+}/Pb}$

$=0.771 V-(-0.126 )=0.897 V$

The over all reaction : 2 × [1] + [2]

$2Fe^{3+}(aq)+Pb(s)\rightarrow 2Fe^{2+}(s)+Pb^{2+}(aq)$

The standard cell potential of the reaction is 0,.897 Volts.

7 0

4 years ago