Eaterfication experiment

When a warm soda can is placed in an ice chest full of ice, the soda can _______ heat and the ice _______ heat until they are bo

Svetach [21]

D because as the soda can starts to cool off the ice in turn begins to melt because the heat from the soda is being distributed on the ice and vice versa.

5 0

2 years ago

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.

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

Hence,

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

8 0

3 years ago

How many moles are in 2.0 grams of Na?

lisov135 [29]

Answer:

46

Explanation:

Sodium metal has a molar mass of

22.99

3 0

2 years ago

a flight stimulator that helps astronauts prepare for a shuttle launch is an example of a (an) ______.

Dimas [21]

A flight simulator that helps astronauts prepare for a shuttle launch is an example of a Scientific Model.

3 0

3 years ago

Read 2 more answers

What is the oxidizing agent in the reaction 2MnO4-(aq) 10I-(aq) 16H (aq) ---> 2Mn2 (aq) 5I2(aq) 8H2O(l)

faltersainse [42]

Correct Question: what is the oxidizing agent in the reaction.

2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Answer: MnO4-is the oxidizing agent

Explanation:

In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)

Oxidizing agent oxidizes other molecules while the themselves get reduced.

oxidizing agents give away Oxygen to other compounds.

MnO4-is the oxidizing agent because

On the reactants side

Oxidation number of Mn in 2MnO4- is +7

Oxidation number of Cl- is -1

On the products side

Oxidation number of Mn is +2

While oxidation number of Cl is zero

Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2

4 0

3 years ago

Eaterfication experiment ​

Eaterfication experiment