Chemistry, 24.03.2020 05:37 aROSSconpollo

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (aq)(aq)(g)(l) (aq)(aq) Answer the following questions about this cell. a. Write a balanced equation for the half-reaction that happens at the cathode. b. Write a balanced equation for the half-reaction that happens at the anode. c. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. d. Do you have enough information to calculate the cell voltage under standard conditions

Answers: 1

Show answers

Answers

Answer from: pineapplepizaaaaa

The question is incomplete, here is the complete question:

A chemist designs a galvanic cell that uses these two half-reactions:

Half-reaction Standard reduction potential

$NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l)$ 0.96 V

$Fe^{3+}(aq.)+e^-\rightarrow Fe^{2+}(aq.)$ 0.77 V

Answer the following questions about this cell.

a. Write a balanced equation for the half-reaction that happens at the cathode.

b. Write a balanced equation for the half-reaction that happens at the anode.

c. Write a balanced equation for the overall reaction that powers the cell.Be sure the reaction is spontaneous as written.

d. Do you have enough information to calculate the cell voltage under standard conditions

For a: The half reaction at cathode is $NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l)$

For b: The half reaction at anode is $Fe^{3+}(aq.)+e^-\rightarrow Fe^{2+}(aq.)$

For c: The overall reaction is $NO_3^-(aq.)+4H^+(aq.)+Fe^{3+}(aq.)\rightarrow NO(g)+2H_2O(l)+Fe^{2+}(aq.)$

For d: It is not possible to calculate cell voltage under given conditions.

Explanation:

The given chemical equation follows:

$NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l)$ E^o=0.96V

$Fe^{3+}(aq.)+e^-\rightarrow Fe^{2+}(aq.)$ E^o=0.77V

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

Here, nitrogen trioxide is getting reduced easily and will undergo reduction reaction and Iron will undergo oxidation reaction.

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

For a:

The half reaction occurring at cathode follows:

$NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l)$

For b:

The half reaction occurring at anode follows:

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

For c:

Oxidation half reaction: $Fe^{3+}(aq.)+e^-\rightarrow Fe^{2+}(aq.)$ ( × 3)

Reduction half reaction: $NO_3^-(aq.)+4H^+(aq.)+3e^-\rightarrow NO(g)+2H_2O(l)$

Overall reaction: $NO_3^-(aq.)+4H^+(aq.)+Fe^{3+}(aq.)\rightarrow NO(g)+2H_2O(l)+Fe^{2+}(aq.)$

For d:

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[\text{Products}]}{[\text{Reactants}]}$

For calculating the cell voltage, we require the concentrations of gaseous and aqueous products and reactants.

Hence, it is not possible to calculate cell voltage under given conditions.

Answer from: Quest

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and conduct electricity easily while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

Answer from: Quest

Wagons/motorized vehicles

Another question on Chemistry

Chemistry, 21.06.2019 23:30

Two atoms interact with each other as shown by the equation. complete the equation by filling in the missing parts. 1 2 3 4 5 h he li

Answers: 2

Answer

Chemistry, 22.06.2019 04:00

Two nitro no2 groups are chemically bonded to a patch of surface. they can't move to another location on the surface, but they can rotate (see sketch at right). it turns out that the amount of rotational kinetic energy each no2 group can have is required to be a multiple of ε, where =ε×1.010−24 j. in other words, each no2 group could have ε of rotational kinetic energy, or 2ε, or 3ε, and so forth — but it cannot have just any old amount of rotational kinetic energy. suppose the total rotational kinetic energy in this system is initially known to be 32ε. then, some heat is removed from the system, and the total rotational kinetic energy falls to 18ε. calculate the change in entropy. round your answer to 3 significant digits, and be sure it has the correct unit symbol.

Answers: 2

Answer

Chemistry, 22.06.2019 16:00

How could a student test the effect of removing heat from a gas that is stored in a sealed container? what must occur in order for matter to change states?

Answers: 2

Answer

Chemistry, 22.06.2019 21:00

What is the chemical formula for the compound formed between sodium and flour one

Answers: 1

Answer

You know the right answer?

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reducti...

Questions