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  • A concentration cell is made up of two <img alt="\mathrm{Ag} / \mathrm{Ag}^{+}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BAg%7D%20/%20%5Cmathrm%7BAg%7D%5E%7B&plus;%7

A concentration cell is made up of two \mathrm{Ag} / \mathrm{Ag}^{+} electrodes, where one electrode has \left[\mathrm{Ag}^{+}\right]=0.01 \mathrm{M} and the other has \left[A g^{+}\right]=0.001 M. What is the cell potential at 25^{\circ} \mathrm{C} if the cell is connected with a salt bridge and the standard potential \mathrm{Ag} / \mathrm{Ag}^{+} of electrode is 0.80V with respect to S.H.E. ?

Option: 1

0.050 V


Option: 2

0.100 V


Option: 3

0.150 V


Option: 4

0.200 V


Answers (1)

best_answer

The cell's balanced redox equation is:

A g^{+}(a q)+e^{-} \rightarrow A g(s)

This half-reaction's standard reduction potential is:

E^{\circ}\left(\mathrm{Ag}^{+} / \mathrm{Ag}\right)=0.80 \mathrm{~V}

At 25°C, the Nernst equation becomes:

\text { Ecell }=0.80 \mathrm{~V}-(0.0592 \mathrm{~V} / 1) \log (Q)

where Q=A g^{+} / A g^{+}

Plugging in the values, we get:

\begin{aligned} & \text { Ecell }=0.80 \mathrm{~V}-(0.0592 \mathrm{~V}) \log (0.001 / 0.01) \\ & \text { Ecell }=0.80 \mathrm{~V}-0.0592 \mathrm{~V} \\ & \text { Ecell }=0.7408 \mathrm{~V} \end{aligned}

Since the higher concentration of silver ions will lose electrons and the lower concentration will gain electrons, the cell potential will be the difference in potentials of the two electrodes:

\text { Ecell }=(0.01-0.001) V=0.009 \mathrm{~V}

The correct option is (1) 0.050 V. 

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Riya

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