In the cell, Zn ( s ) ∣ Zn + 2 ( C 1 ) ∣∣ Cu + 2 ( C 2 ) ∣ Cu ( s ) ; E ce ll − E ce ll ∘ = 0.0591 volt. The ratio ( C 1 / C 2 ) at 298 K will be:

Question

In the cell, Zn ( s ) ​ ∣ Zn + 2 ( C 1 ​ ) ∣∣ Cu + 2 ( C 2 ​ ) ∣ Cu ( s ) ; E ce ll ​ − E ce ll ∘ ​ = 0.0591 volt. The ratio ( C 1 ​ / C 2 ​ ) at 298 K will be:

TestHub · Accepted Answer

The Nernst equation for the given cell is $E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.0591}{n} lo g \frac{C _{1}}{C _{2}}$ .

Given $E_{ce ll} - E_{ce ll}^{\circ} = 0.0591$ and $n = 2$ .

So, $0.0591 = - \frac{0.0591}{2} lo g \frac{C _{1}}{C _{2}}$ .

This simplifies to $- 2 = lo g \frac{C _{1}}{C _{2}}$ .

Therefore, $\frac{C _{1}}{C _{2}} = 1 0^{- 2}$ .

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