MHT CET · Chemistry · Electrochemistry
What is the relation between cell constant, conductivity and electrical resistance?
- A \(\mathrm{k}=\frac{\mathrm{R}}{\mathrm{b}}\)
- B \(\mathrm{k}=\frac{\mathrm{b}}{\mathrm{R}}\)
- C \(\mathrm{k}=\frac{1}{\mathrm{R} . \mathrm{b}}\)
- D \(\mathrm{k}=\mathrm{R} \cdot \mathrm{b}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{k}=\frac{\mathrm{b}}{\mathrm{R}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_{\mathrm{cell}}^{0}=0.463 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Cu}}^{0}=0.337 \mathrm{~V}\)
\(\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}}^{0}\)
\(\therefore \mathrm{E}_{\mathrm{Ag}}^{0}=\mathrm{E}_{\mathrm{cell}}^{0}+\mathrm{E}_{\mathrm{Cu}}^{0}=0.463+0.337=0.800 \mathrm{~V}\)
\(\mathrm{E}_{\mathrm{cell}}^{0}=\mathrm{E}_{\mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}}^{0}\)
\(\therefore \mathrm{E}_{\mathrm{Ag}}^{0}=\mathrm{E}_{\mathrm{cell}}^{0}+\mathrm{E}_{\mathrm{Cu}}^{0}=0.463+0.337=0.800 \mathrm{~V}\)
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