Which element is oxidized in the following reaction: cuo(s) + h2 (g) cu(s) + h2o (l)
Oxide + hydrogen is the element that is oxidized.
Answer : The oxidized element is, hydrogen. Explanation : Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously. Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place. Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place. The balanced redox reaction is : The half oxidation-reduction reactions are: Oxidation reaction : Reduction reaction : In this reaction, the oxidation state of ‘H’ changes from (0) to (+1) that means ‘H’ lost 1 electron and it shows oxidation reaction and the oxidation state of ‘Cu’ changes from (+2) to (0) that means ‘Cu’ gain 1 electron and it shows reduction reaction. Thus, ‘H’ is oxidized and ‘Cu’ is reduced.
you are a god Explanation:
13) phosphoric acid + potassium hydroxide H3PO4 + 3 KOH > K3PO4 + 3H2O H3PO4 ( acid ) KOH ( base ) K3PO4 ( salt ) Therefore fore K3PO4 is salt ( tripotassium phosfate ) 14) Al(ClO4)3 Al⁺³ ClO₄⁻ in chlorine = x + ( – 2 * 4 ) = -1 x – 8 +1 = 0 x = 8 – 1 x = +7 in Al(ClO4)3 chlorine oxidation state + 7 15) CuO(s) + H2 (g) Cu(s) + H2O (l) The oxidation state of Cu in CuO = +2 The oxidation state of H in H2 = 0 The oxidation state of Cu in Cu = 0 The oxidation state of H in H2O = +1 That in the above reaction , The oxidation state of Cu is decreased from +2 to 0 , so Cu⁺² is reduced , The oxidation state of H is decreased from 0 to +1 , So H2 is oxidized . Hope that helps!
Start by determining the oxidation numbers of each element on both sides of the equation CuO(s) + H2 (g) → Cu(s) + H2O (l) Let’s start with the reactants. In the compound CuO(s), the oxidation number of O is -2 and the oxidation number of Cu is +2. (Remember, oxygen’s oxidation number is -2 unless it’s in a peroxide) In H2(g) the oxidation number of H is 0 because it is in its elemental form. Now onto the products The oxidation number of Cu(s) is 0 because it is in its elemental form In H2O(l) the oxidation number of oxygen is -2, and the oxidation number of H is +1. Now let’s summarize the change in the oxidation numbers of each element. Cu: started as +2 and became 0 O: started as -2 and became -2 (didn’t change) H: started as 0 and became +1 To determine which element was oxidized, we look at which element lost electrons, and thus became more positive. So, we can determine that H was oxidized in this reaction, because it started with the oxidation number 0 and became +1.
Answer 6
Oxide + hydrogen is the element that is oxidized.
Answer 7
Answer : The oxidized element is, hydrogen. Explanation : Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously. Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place. Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place. The balanced redox reaction is : The half oxidation-reduction reactions are: Oxidation reaction : Reduction reaction : In this reaction, the oxidation state of ‘H’ changes from (0) to (+1) that means ‘H’ lost 1 electron and it shows oxidation reaction and the oxidation state of ‘Cu’ changes from (+2) to (0) that means ‘Cu’ gain 1 electron and it shows reduction reaction. Thus, ‘H’ is oxidized and ‘Cu’ is reduced.
you are a god Explanation:
13) phosphoric acid + potassium hydroxide H3PO4 + 3 KOH > K3PO4 + 3H2O H3PO4 ( acid ) KOH ( base ) K3PO4 ( salt ) Therefore fore K3PO4 is salt ( tripotassium phosfate ) 14) Al(ClO4)3 Al⁺³ ClO₄⁻ in chlorine = x + ( – 2 * 4 ) = -1 x – 8 +1 = 0 x = 8 – 1 x = +7 in Al(ClO4)3 chlorine oxidation state + 7 15) CuO(s) + H2 (g) Cu(s) + H2O (l) The oxidation state of Cu in CuO = +2 The oxidation state of H in H2 = 0 The oxidation state of Cu in Cu = 0 The oxidation state of H in H2O = +1 That in the above reaction , The oxidation state of Cu is decreased from +2 to 0 , so Cu⁺² is reduced , The oxidation state of H is decreased from 0 to +1 , So H2 is oxidized . Hope that helps!
0
Start by determining the oxidation numbers of each element on both sides of the equation CuO(s) + H2 (g) → Cu(s) + H2O (l) Let’s start with the reactants. In the compound CuO(s), the oxidation number of O is -2 and the oxidation number of Cu is +2. (Remember, oxygen’s oxidation number is -2 unless it’s in a peroxide) In H2(g) the oxidation number of H is 0 because it is in its elemental form. Now onto the products The oxidation number of Cu(s) is 0 because it is in its elemental form In H2O(l) the oxidation number of oxygen is -2, and the oxidation number of H is +1. Now let’s summarize the change in the oxidation numbers of each element. Cu: started as +2 and became 0 O: started as -2 and became -2 (didn’t change) H: started as 0 and became +1 To determine which element was oxidized, we look at which element lost electrons, and thus became more positive. So, we can determine that H was oxidized in this reaction, because it started with the oxidation number 0 and became +1.
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