Consider a reaction spontaneously proceeding in the forward direction. Use this

Question

Consider a reaction spontaneously proceeding in the forward direction.

Use this equation to answer the following questions: ?G=?G?+RTlnQ.

Part H

As a reaction proceeds in the forward direction, the value of Q  Please Chooseincreases. decreases.

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Part I

Therefore, the nonstandard value of ?G becomes Please Choosemore negative. more positive.

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Part J

A voltaic cell reaction becomes Please Chooseless spontaneous more spontaneous  as a reaction proceeds.

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Part K - Recall: ?G=?G?+RTlnQ

Does the change in Q (you described in Parts G and H) cause the value of nonstandard ?G to become more positive or less positive than ?Go? (Remember that ?G should be negative for the spontaneous process.)

The bigger the Q  

Does the change in  (you described in Parts G and H) cause the value of nonstandard  to become more positive or less positive than ? (Remember that  should be negative for the spontaneous process.)

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For the following questions, refer to the relationship between nonstandard ?G and Ecell.

?G=?nFEcell

Part L - For a working voltaic cell, a spontaneous reaction is occurring.

The signs of ?G and Ecell are Please Choose(-) and (-) (+) and (+) (+) and (-) (-) and (+)  , respectively.

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Part M - What are the values of ?G and Ecell when the reaction in the cell has reached a state of equilibrium?

?G and Ecell are Please Chooseboth zero. both greater than one. both less than one. (+) and (+), respectively. (+) and (-), respectively. (-) and (+), respectively. (-) and (-), respectively.

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Part N - Consider how ?G changes as a reaction proceeds toward equilibrium (Parts I-K, above).

The value of Ecell becomes a

as the reaction in the voltaic cell proceeds in the forward direction toward a state of equilibrium.

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Part O - Recall: ?G=?nFEcell

As the value of Ecell becomes a smaller positive value, this results in a ________ value of ?G and therefore a _______ spontaneous reaction.

Please Chooselarger positive, more larger negative, more smaller positive, less smaller negative, less

Explanation / Answer

with, dG = dGo + RTlnQ

Part H : As the reaction proceeds in the forward direction, the value of Q ,

increases

Part I : Therefore the value of nonstandard dG becomes,

more negative (value becomes smaller as we are now substracting larger RTlnQ value from dGo)

Part J : A voltaic cell becomes more,

less spontaneous as reaction proceeds (tries to reach a state of equilibrium)

Part K : dG = dGo + RTlnQ

change in Q causes the value of nonstandard dG to become,

more positive

--

With dG = -nFEcell

Part L : spontaneous voltaic cell

sign of dG = -ve

sign of Ecell = +ve

Part M: at equilibrium

dG = 0

Ecell = 0

Part N : as reaction proceeds towards equilibrium,

The value of Ecell becomes a,

smaller positive value

Part O : As the value of Ecell becomes a smaller positive value, this results in a less smaller negative value of dG and therefore a less spontaneous reaction.

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