In the electrochemical cell ${\rm{Zn|ZnS}}{{\rm{O}}_4}(0.01{\rm{M}})||{\rm{CuS}}{{\rm{O}}_{\rm{4}}}(1.0{\rm{M}}){\rm{|Cu,}}$ the emf of this Daniel cell is ${{\rm{E}}_1}.$ When the concentration of ${\rm{ZnS}}{{\rm{O}}_4}$ is changed to 1.0 M and that of ${\rm{CuS}}{{\rm{O}}_{\rm{4}}}$ changed to 0.01 M, the emf changes to ${{\rm{E}}_2}.$ From the following , which one is the relationship between ${{\rm{E}}_{\rm{1}}}{\rm{ \;and\;}}\,{{\rm{E}}_2}\,?$ (Given, $\frac{{{\rm{RT}}}}{{\rm{F}}} = 0.059$)

${{\rm{E}}_2} = 0 \ne {{\rm{E}}_1}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Application +4 | -1

In the electrochemical cell

\mathrm{Zn|ZnS}{\mathrm{O}_4}(0.01\mathrm{M})||\mathrm{CuS}{\mathrm{O}_\mathrm{4}}(1.0\mathrm{M})\mathrm{|Cu,}

the emf of this Daniel cell is

{\mathrm{E}_1}.

When the concentration of

\mathrm{ZnS}{\mathrm{O}_4}

is changed to 1.0 M and that of

\mathrm{CuS}{\mathrm{O}_\mathrm{4}}

changed to 0.01 M, the emf changes to

{\mathrm{E}_2}.

From the following , which one is the relationship between

{\mathrm{E}_\mathrm{1}}\mathrm{ \;and\;}\,{\mathrm{E}_2}\,?

(Given,

\frac{{\mathrm{RT}}}{\mathrm{F}} = 0.059

)

View answer and explanation

Correct option: D

{\mathrm{E}_1} > {\mathrm{E}_2}

Option D (

{\mathrm{E}_1} > {\mathrm{E}_2}

) is correct because, according to the Nernst equation, increasing the concentration of the reactant and decreasing the concentration of the product will increase the cell potential.

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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