MHT CET 2024 2nd May Morning Shift | Electrochemistry Question 48 | Chemistry

A conductivity cell containing $$0.001 \mathrm{~M} \mathrm{~AgNO}_3$$ solution develops resistance $$6530 \mathrm{ohm}$$ at $$25^{\circ} \mathrm{C}$$. Calculate the electrical conductivity of solution at same temperature if the cell constant is $$0.653 \mathrm{~cm}^{-1}$$.

$$1.3 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$$

$$1.5 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$$

$$1.7 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$$

$$1.0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$$

MHT CET 2023 14th May Evening Shift

MCQ (Single Correct Answer)

-0

Calculate $$\mathrm{E}_{\text {cell }}^0$$ for $$\mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{1M})}^{++}\right|\left|\mathrm{Ag}_{(\mathrm{1M})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}$$.

$$\left[\mathrm{E}_{\mathrm{Cd}}^0=-0.403 \mathrm{~V} ; \mathrm{E}_{\mathrm{Ag}}^0=0.799 \mathrm{~V}\right.\text {]}$$

$$1.202 \mathrm{~V}$$

$$-1.202 \mathrm{~V}$$

$$0.396 \mathrm{~V}$$

$$-0.396 \mathrm{~V}$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

Which from following expressions is used to find the cell potential of $$\mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{aq})}^{++}\right|\left|\mathrm{Cu}_{(\mathrm{aq})}^{+}\right| \mathrm{Cu}_{(\mathrm{s})}$$ cell at $$25^{\circ} \mathrm{C}$$ ?

$$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o-0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]}$$

$$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o+0.0296 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]}$$

$$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^o-0.0592 \log \frac{\left[\mathrm{Cu}^{++}\right]}{\left[\mathrm{Cd}^{++}\right]}$$

$$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}+0.0592 \log \frac{\left[\mathrm{Cu}^{++}\right]}{\left[\mathrm{Cd}^{++}\right]}$$

PREVIOUS NEXT

MHT CET Subjects