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 ............ $V$.

$(i)$ Copper metal dissolves in $1 \ M$ silver nitrate solution and crystals of silver metal get deposited.
$(ii)$ Silver metal does not react with $1 \ M$ zinc nitrate solution.
$(iii)$ Zinc metal dissolves in $1 \ M$ copper sulphate solution and copper metal gets deposited.
Hence,the order of decreasing strength of the three metals as reducing agents will be:

The $E^{0}_{Red}$ values for $P, Q, R$ and $S$ are $-2.90 \, V, +0.34 \, V, +1.20 \, V$ and $-0.76 \, V$ respectively. The decreasing order of their reactivity is:

Electrode potential data are given below :
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:

Using the standard reduction potentials of the electrodes $Li$,$Zn$,$Mg$,and $Ni$ as $-3.05 \ V$,$-0.76 \ V$,$-2.36 \ V$,and $-0.25 \ V$ respectively,identify the correct statement.

For $Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}$,$E^{0} = -2.37 \, V$ and for $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$,$E^{0} = +0.33 \, V$. The $emf$ of the cell $(E^{0}_{Cell})$ is .......... $V$.