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  • Consider a cell reaction <img alt="\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq})" src="https://entrancecorner.oncodec

Consider a cell reaction \mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) having a standard cell potential of 1.10 \mathrm{~V}$ at $298 \mathrm{~K}. The concentration of \mathrm{Fe}^{3+} ions in the cell is 0.1 \mathrm{M}. What is the concentration of \mathrm{Zn}^{2+} ions required in the cell so that the cell potential is 1.20 \mathrm{~V}\; \text{at} \; 298 \mathrm{~K} ? (\text{Given}: R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, F=96485 \mathrm{Cmol}^{-1} )

 

Option: 1

0.032 M


Option: 2

0.0325 M


Option: 3

0.033 M


Option: 4

0.0335 M


Answers (1)

best_answer

The Nernst equation for a galvanic cell is given as:

\mathrm{E}_{\text {cell }}=\mathrm{E}^{\ominus} c e l l-\frac{R T}{n F} \ln Q

Where, 

\mathrm{E}_{\text {cell }} is the cell potential
\mathrm{E}^{\ominus}\text{cell} is the standard cell potential
\mathrm{R} is the gas constant
\mathrm{T} is the temperature in Kelvin
\mathrm{n} is the number of electrons transferred in the cell reaction
\mathrm{F} is the Faraday constant

\mathrm{Q} is the reaction quotient, given by \frac{\left[\mathrm{Ze}^{3+}\right]}{\left[\mathrm{Fe}^{3+}\right]} for the given cell.
Since the standard cell potential is given as 1.10 \mathrm{~V} \; \text{at}\; 298 \mathrm{~K}, we have:
\mathrm{E}^{\Theta} \text { cell }=1.10 \mathrm{~V}
To find the concentration of \mathrm{Zn}^{2+} ions required to obtain a cell potential of 1.20 \mathrm{~V} \; \text{at }298 K, we can use the Nernst equation as follows:

\\1.20=1.10-\frac{(8.314)(298)}{2(96485)} \ln \frac{\left[\mathrm{Zn}^{2+}\right]}{0.1} \\\\ \ln \frac{\left[\mathrm{Zn}^{2+}\right]}{0.1}=-\frac{2(1.20-1.10)(96485)}{(8.314)(298)}\\\\\left[\mathrm{Zn}^{2+}\right]=0.033 \mathrm{M}

Therefore, the concentration of \mathrm{Zn}^{2+} ions required in the cell so that the cell potential is 1.20 \mathrm{~V} \; \text{at} \; 298 \mathrm{~K} \; \text{is} \; 0.033\; \mathrm{M}. Hence, the correct answer is 3.

Posted by

SANGALDEEP SINGH

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