What event occurs when the

What substance is a combination of different atoms

MrMuchimi

I'm not suree.. is his science?

7 0

4 years ago

4. Assuming the reaction below is at equilibrium, which of the following changes will drive the reaction to the left? C(s) + O2(

shepuryov [24]

Answer:

Increasing temperature

Explanation:

$C(s) + O_2(g) \leftrightharpoons CO_2(g)$

Enthalpy of the reaction = -393.5 kJ/mol

Negative sign implies that reaction is exothermic.

Effect of change in reaction condition is explained by Le chateliers principle.

According to Le chateliers principle, if the reaction conditions of a reversible reaction in a state of dynamic equilibirum is changed, the reaction will move in a direction to counteract the change.

1. Increasing the temperature

Forward reaction is exothermic that means temperature increases in forward direction. Backward reaction will be endothermic and so there is decrease in temperature in backward direction or in left direction.

On increasing temperature, reaction will be move in direction to counteract the increased temperature, therefore reaction will move in left direction.

2. Adding O2

If O2 is added, then reaction will move in a direction in which its get consumed. So, reaction will move in forward direction or in right direction.

3. Removing C (s)

Le Chatelier's principle does not apply on solids, so removal of C(s) does not affect the equilibrium.

4 0

4 years ago

Read 2 more answers

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$