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(a) How do the relative magnitudes of Q and K relate to the signs of Δ³ÒandEcell ? Explain.

(b) Can a cell do work when Q/K> 1or Q/K< 1? Explain.

Short Answer

Expert verified

a. The dependency of the sign of Ecell  and  Δ³Ò is given below.

  • When, QK=1, â¶Ä‰Ecell=0  and  Δ³Ò=0
  • When, QK&±ô³Ù;1, â¶Ä‰â¶Ä‰l²ÔQK<0. Therefore, Ecell>0   and   Δ³Ò<0.Thus the reaction proceeds spontaneously.
  • When, QK&²µ³Ù;1, â¶Ä‰â¶Ä‰l²ÔQK>0. Therefore,Ecell<0   and   Δ³Ò>0. Thus the reaction proceeds non- spontaneously.

b. A cell is spontaneous when QK is less than 1.

Step by step solution

01

Sign of   Ecell  and  ΔG

When a redox reaction in a cell is in equilibrium under standard conditions, the equilibrium constant is denoted by K. The equilibrium constant under non-standard conditions is denoted by Q.

The Nernst equation is given below.

Ecell=Ecello-RTnF±ô²Ô²Ï â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰(1)

We also know,

E0cell = â¶Ä‰RTnF±ô²Ô°­â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰(2)

Substituting the value of E0cell in equation (1), we get:

role="math" localid="1659591790741" Ecell= â¶Ä‰â¶Ä‰RTnF±ô²Ô°­â€‰â¶Ä‰-RTnF±ô²Ô²Ï â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰=-RTnFlnQK â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰(3)

Now, we will see three conditions for predicting the signs for Ecell  and  Δ³Ò.

  • When, QK=1, â¶Ä‰Ecell=0  and  Δ³Ò=0
  • When, QK&±ô³Ù;1, â¶Ä‰â¶Ä‰l²ÔQK<0. Therefore,Ecell>0   and   Δ³Ò<0.Thus the reaction proceeds spontaneously.
  • When, QK&²µ³Ù;1, â¶Ä‰â¶Ä‰l²ÔQK>0. Therefore, Ecell<0   and   Δ³Ò>0.Thus the reaction proceeds non- spontaneously.
02

Spontaneity of the cell

When QK&±ô³Ù;1, â¶Ä‰â¶Ä‰l²ÔQK<0.

Consequently, Ecell>0   and   Δ³Ò<0. Thus, a cell is spontaneous when QKis less than 1.

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Most popular questions from this chapter

Comparing the standard electrode potentials (Eo)of the Group 1A(1)metalsLi, Na, and Kwith the negative of their first ionization energies reveals a discrepancy:

Ionization process reversed:M+(g) +e-⇌M(g) ( - IE)

Electrode reaction:M+(aq) +e-⇌M(s) (Eo)

Note that the electrode potentials do not decrease smoothly down the group, as the ionization energies do. You might expect that if it is more difficult to remove an electron from an atom to form a gaseous ion (largerIE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smallerEo), yetLi+(aq)is more difficult to reduce thanNa+(aq). Applying Hess’s law, use an approach similar to that for a Born-Haber cycle to break down the process occurring at the electrode into three steps and label the energy involved in each step. How can you account for the discrepancy?

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(a)role="math" localid="1663418747793" Cl2(g) + Fe2 +(aq) Cl-(aq) + Fe3 +(aq)

(b)role="math" localid="1663418761168" Mn2 +(aq) + Co3 +(aq) MnO2(s) + Co2 +(aq)[acidic]

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