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

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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.

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Answer from: Quest

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