What is the value of $n$ in the Daniell cell? Why?
(N/A) In the Daniell cell,the overall cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.
Here,$Zn$ is oxidized to $Zn^{2+}$ by losing $2$ electrons,and $Cu^{2+}$ is reduced to $Cu$ by gaining $2$ electrons.
Since the number of electrons transferred in the balanced redox reaction is $2$,the value of $n$ is $2$.
Here,$Zn$ is oxidized to $Zn^{2+}$ by losing $2$ electrons,and $Cu^{2+}$ is reduced to $Cu$ by gaining $2$ electrons.
Since the number of electrons transferred in the balanced redox reaction is $2$,the value of $n$ is $2$.
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