Categories > electrochemistry > Electrode potential > The emf of a Daniel cell at 298 K is ${{E}_{1}}Zn|\underset{(0.01\,M)}{\mathop{ZnS{{O}_{4}}}}\,||\underset{(1.0\,M)}{\mathop{CuS{{O}_{4}}}}\,|Cu$. When the concentration of $ZnS{{O}_{4}}$ is 1.0 M and that of $CuS{{O}_{4}}$ is 0.01 M, the emf changed to ${{E}_{2}}$. What is the relationship between ${{E}_{{}}}$ and ${{E}_{2}}$

electrochemistry

Question: The emf of a Daniel cell at 298 K is ${{E}_{1}}Zn|\underset{(0.01\,M)}{\mathop{ZnS{{O}_{4}}}}\,||\underset{(1.0\,M)}{\mathop{CuS{{O}_{4}}}}\,|Cu$. When the concentration of $ZnS{{O}_{4}}$ is 1.0 M and that of $CuS{{O}_{4}}$ is 0.01 M, the emf changed to ${{E}_{2}}$. What is the relationship between ${{E}_{{}}}$ and ${{E}_{2}}$


1) ${{E}_{2}}=0\ne {{E}_{1}}$
2) ${{E}_{1}}>{{E}_{2}}$
3) ${{E}_{1}}<{{E}_{2}}$
4) ${{E}_{1}}={{E}_{2}}$
Solution: Explanation: No Explanation
Electrode potential Ecell Nernst equation and ECS

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