determine whether each of the following equations is a redox reaction or not
C6H12O6(s)→CO2(g)+C2H5OH(l)
NaOH(aq)+Al(s)+H2O(l)→H2(g)+NaAlO2(s)
SO2(g)+H2O(l)→H2SO3(aq)
2NaOH(aq)+Cl2(g)→NaCl(aq)+H2O(l)+NaClO(aq)
Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
Fe3O4(s)+CO(g)→3FeO(s)+CO2(g)
2Fe(OH)3(s)→Fe2O3(s)+3H2O(l)
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6 years ago
Answered By Leonardo F
In order for us to have a redox equation, one chemical element must have their oxidation number increased, while another must have their oxidation state decreased. If the same chemical element both increases and decreases its oxidation number, then the equation won't be a redox one. Let's analyse them one by one:
1) C6H12O6(s)→CO2(g)+C2H5OH(l)
The carbon in C6H12O6 has an oxidation number of 0, because the oxidation states of H and O are, respectively, +1 and -2. The carbon in CO2 has an oxidation state of +4 and, in C2H5OH, -2. In this case, carbon decreases and increases its oxidation state, so this is NOT a redox equation.
2) NaOH(aq)+Al(s)+H2O(l)→H2(g)+NaAlO2(s)
The aluminum in Al(s) has oxidation number 0. In NaAlO2, +3, because Na is +1 and O is -2. Hence, the aluminum is oxidating. The hydrogen in H2O has oxidation number of +1. In H2(g), 0. So, the hyrogen is reducing. This IS a redox equation.
3) SO2(g)+H2O(l)→H2SO3(aq)
The sulfur in SO2 has an oxidation number of +4. In H2SO3, also -4, because H is +1 and O is -2. The oxygen has oxidation number of -2 and H +1. None of the elements is changing their oxidation states. So, this is NOT a redox equation.
4) 2NaOH(aq)+Cl2(g)→NaCl(aq)+H2O(l)+NaClO(aq)
The chlorine in Cl2 has oxidation number 0. In NaCl, -1 and, in NaClO, +1. The same chemical element is increasing and decreasing its oxidation state. This is NOT a redox equation.
5) Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)
The carbon is Na2CO3 has an oxidation number of +4, because Na is +1 and O is -2. In CO2, the carbon is also +4. Hence, none of the chemical elements is changing their oxidation state. This is NOT a redox equation.
6) C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
The oxygen in O2 has an oxidation state of 0. In CO2 or in H2O, -2. Hence, it's reducing. The carbon in C6H12O6 has oxidation number of 0. In CO2, +4. So, the carbon is oxidizing. This IS a redox equation.
7) Fe3O4(s)+CO(g)→3FeO(s)+CO2(g)
Let's calculate the oxidation number of Fe in Fe3O4. The oxidation state of O is -2. Hence:
3x + 4(-2) = 0
x = 8/3
The oxidation state of Fe is 8/3 in Fe3O4. In FeO, the oxidation number of Fe is +2. Hence, Fe is reducing. The carbon in CO is +2. In CO2, it's +4. So, the carbon is oxidizing. This IS a redox equation.
6 years ago
Answered By Leonardo F
In order for us to have a redox equation, one chemical element must have their oxidation number increased, while another must have their oxidation state decreased. If the same chemical element both increases and decreases its oxidation number, then the equation won't be a redox one. Let's analyse them one by one:
1) C6H12O6(s)→CO2(g)+C2H5OH(l)
The carbon in C6H12O6 has an oxidation number of 0, because the oxidation states of H and O are, respectively, +1 and -2. The carbon in CO2 has an oxidation state of +4 and, in C2H5OH, -2. In this case, carbon decreases and increases its oxidation state, so this is NOT a redox equation.
2) NaOH(aq)+Al(s)+H2O(l)→H2(g)+NaAlO2(s)
The aluminum in Al(s) has oxidation number 0. In NaAlO2, +3, because Na is +1 and O is -2. Hence, the aluminum is oxidating. The hydrogen in H2O has oxidation number of +1. In H2(g), 0. So, the hyrogen is reducing. This IS a redox equation.
3) SO2(g)+H2O(l)→H2SO3(aq)
The sulfur in SO2 has an oxidation number of +4. In H2SO3, also -4, because H is +1 and O is -2. The oxygen has oxidation number of -2 and H +1. None of the elements is changing their oxidation states. So, this is NOT a redox equation.
4) 2NaOH(aq)+Cl2(g)→NaCl(aq)+H2O(l)+NaClO(aq)
The chlorine in Cl2 has oxidation number 0. In NaCl, -1 and, in NaClO, +1. The same chemical element is increasing and decreasing its oxidation state. This is NOT a redox equation.
5) Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)
The carbon is Na2CO3 has an oxidation number of +4, because Na is +1 and O is -2. In CO2, the carbon is also +4. Hence, none of the chemical elements is changing their oxidation state. This is NOT a redox equation.
6) C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
The oxygen in O2 has an oxidation state of 0. In CO2 or in H2O, -2. Hence, it's reducing. The carbon in C6H12O6 has oxidation number of 0. In CO2, +4. So, the carbon is oxidizing. This IS a redox equation.
7) Fe3O4(s)+CO(g)→3FeO(s)+CO2(g)
Let's calculate the oxidation number of Fe in Fe3O4. The oxidation state of O is -2. Hence:
3x + 4(-2) = 0
x = 8/3
The oxidation state of Fe is 8/3 in Fe3O4. In FeO, the oxidation number of Fe is +2. Hence, Fe is reducing. The carbon in CO is +2. In CO2, it's +4. So, the carbon is oxidizing. This IS a redox equation.