Q21.52P

Question

(a) How do the relative magnitudes of Q and K relate to the signs of $ΔG$ and $E_{cell}$ ? Explain.

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

Step-by-Step Solution

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Answer

a. The dependency of the sign of $E_{cell} and ΔG$ is given below.

When, $\frac{Q}{K} = 1, E_{cell} =0 and ΔG=0$
When, $\frac{Q}{K} <1, ln \frac{Q}{K} <0.$ Therefore, $E_{cell} >0 and ΔG<0.$ Thus the reaction proceeds spontaneously.
When, $\frac{Q}{K} >1, ln \frac{Q}{K} >0.$ Therefore, $E_{cell} <0 and ΔG>0.$ Thus the reaction proceeds non- spontaneously.

b. A cell is spontaneous when $\frac{Q}{K}$ is less than 1.

1Step 1: Sign of E cell    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.

$E_{cell} {=E}_{cell}^{o} - \frac{RT}{nF} lnQ (1)$

We also know,

${E^{0}}_{cell} = \frac{RT}{nF} lnK (2)$

Substituting the value of ${E^{0}}_{cell}$ in equation (1), we get:

$\begin{matrix} E_{cell} = \frac{RT}{nF} lnK - \frac{RT}{nF} lnQ \\ = - \frac{RT}{nF} ln \frac{Q}{K} (3) \end{matrix}$

Now, we will see three conditions for predicting the signs for $E_{cell} and ΔG$ .

When, $\frac{Q}{K} = 1, E_{cell} =0 and ΔG=0$
When, $\frac{Q}{K} <1, ln \frac{Q}{K} <0.$ Therefore, $E_{cell} >0 and ΔG<0.$ Thus the reaction proceeds spontaneously.
When, $\frac{Q}{K} >1, ln \frac{Q}{K} >0.$ Therefore, $E_{cell} <0 and ΔG>0.$ Thus the reaction proceeds non- spontaneously.

2Step 2: Spontaneity of the cell

When $\frac{Q}{K} <1, ln \frac{Q}{K} <0.$

Consequently, $E_{cell} >0 and ΔG<0.$ Thus, a cell is spontaneous when $\frac{Q}{K}$ is less than 1.

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Q21.53P

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