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ValentinkaMS [17]
3 years ago
10

Consider the following half reaction in which nitrate is reduced to nitrite: NO3– 2H 2e– → NO2– H2O. The reaction has a ΔG°' val

ue of –81 kJ/mol. The negative value of ΔG°' results from the very __________ E°' value of the redox couple NO3–/NO2–, and it reflects the strong tendency of nitrate to __________ electrons. Choose one:A. positive / donateB. negative/ acceptC. negative / donateD. positive / accept
Chemistry
1 answer:
drek231 [11]3 years ago
6 0

Answer:

D. positive/accept

Explanation:

a) The standard gibbs free energy (ΔG⁰) is related to the standard cell potential (E⁰) as follows:

\Delta G^{0}= -nFE^{0}-----(1)

where n = number of electrons

F = faraday constant

The given half  reaction reaction has ΔG⁰ = -81 kJ/mol. Based on equation (1) a negative value for the free energy can be obtained only if E⁰ is positive.

Thus, the negative value of ΔG°' results from the very positive value of E⁰

b) The value of E⁰ is the standard reduction potential which reflects the tendency of the system to accept electrons and get reduced. Higher (more positive) the value, greater will be the tendency to accept electrons.

Therefore, the positive E⁰ value of the redox couple NO3–/NO2– reflects the strong tendency of nitrate to <u>accept </u>electrons.

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Answer:

See detailed explanation.

Explanation:

Hello!

In this case, for the described chemical reaction, we can proceed as follows:

A) For the complete chemical reaction we note down every reacted and produced species as well as the proper balancing process:

2NaBr(aq)+Cl_2(aq)\rightarrow 2NaCl(aq)+Br_2(g)

In which gaseous bromine may give off.

B) The dissociated ionic equation requires the ionization of the aqueous species in ions, expect for chlorine which is not ionized:

2Na^+(aq)+2Br^-(aq)+Cl_2(aq)\rightarrow 2Na^+(aq)+2Cl^-(aq)+Br_2(g)

C) For the net ionic equation we cancel out the sodium ions as they are at both reactants and products:

2Br^-(aq)+Cl_2(aq)\rightarrow +2Cl^-(aq)+Br_2(g)

D) Based on C) we infer that the spectator ions here are the sodium ions.

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When titrating a strong monoprotic acid and koh at 25°c, the ph will be less than 7 at the equivalence point. ph will be greater
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By definition titraion of a monoprotic acid with means that the equivalence point implies netrality of the solution, which is pH = 7.

So, the answer is that pH will be equal to 7 at the equivalence.

Given that the acid is monoprotic and KOH has one OH- radical per molecule of KOH, the titration will require the same number of moles of acid than base to reach the equivalence point, as you can see in this equation, representing the monoprotic acid as HA:

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