If the electrode potential of $Zn^{2+} | Zn$ is $-0.76 \, V$ and the potential of $Cu^{2+} | Cu$ is $0.34 \, V$,what will be the $EMF$ of the cell formed by both electrodes in $V$?
- A$1.10$
- B$0.42$
- C$-1.10$
- D$-0.42$
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Electrode potential of $Zn^{2+}/Zn$ is $-0.76\,V$ and that of $Cu^{2+}/Cu$ is $+0.34\,V$. The $EMF$ of the cell constructed between these two electrodes is ............ $V$.
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View SolutionCalculate the $Zn-Cu$ cell (Daniell cell) potential. Given: $E^o_{Zn^{2+}|Zn} = -0.76 \ V$ and $E^o_{Cu^{2+}|Cu} = 0.34 \ V$. (in $V$)
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