Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The EMF of a concentration cell is found to be -0.059 V at  &nbsp

The EMF of a concentration cell is found to be -0.059 V at  25^{\circ} \mathrm{C} when the concentration of the \mathrm{Fe^{2+}} ions in the anode half-cell is 0.0010 M and that in the cathode half-cell is 0.10 M. What will be the EMF of the cell when the concentration of the ions in the anode half-cell is increased to 0.10 M, and that in the cathode half-cell is decreased to 0.0010 M, assuming that the temperature and the volumes of the half-cells remain constant?

 

Option: 1

0.034 V


Option: 2

0.059 V


Option: 3

-0.034 V


Option: 4

-0.059 V


Answers (1)

best_answer

The Nernst equation for a concentration cell is given by:

E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln (Q)

where E^{\circ} cell is the standard cell potential, R is the gas constant, T is the temperature in kelvins, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient.
When the concentration of the \mathrm{Fe}^{2+} ions in the anode half-cell is 0.0010 \mathrm{M} and that in the cathode half-cell is 0.10 \mathrm{M}, the EMF of the cell is given by:

\\\mathrm{E_{\text {cell }}=E^o_{\text {cell }}-\frac{R T}{n F} \ln Q}\\\\ \mathrm{ E_{\text {cell }}=0-\frac{(0.0592)}{2} \log _{10}\left(\frac{0.10}{0.0010}\right)=-0.059 \mathrm{~V}}

When the concentration of the \mathrm{Fe}^{2+} ions in the anode half-cell is increased to 0.10 M, and that in the cathode half-cell is decreased to 0.0010 M, the EMF of the cell is given by:

\\\mathrm{E_{\text {cell }}=E^{\circ} \text {cell }-\frac{R T}{n F} \ln Q}\\\\ \mathrm{ E_{\text {cell }}=0-\frac{0.0592}{2} \log 10\left(\frac{0.0010}{0.10}\right)=0.034 \mathrm{~V}}

Therefore, the correct answer is (1).

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

Andhra launches free JEE, NEET coaching to over 1 lakh govt school students
anam-khan

JoSAA Counselling 2025 LIVE: BTech registration starts at josaa.nic.in for IIT, NIT, IIIT admissions
anam-khan

Kota: 17-year old student from UP preparing for JEE dies of electrocution in hostel

Latest Question