$Zn(s) + C{u^{2 + }}(0.1\,M) \to Z{n^{2 + }}(1\,M) + Cu(s)$ માટે $E_{cell}^o$ is $1.10\,volt$ હોય તો ${E_{cell}}$ જણાવો. $\left( {2.303\frac{{RT}}{F} = 0.0591} \right)$
\( = 1.10 - 0.0295\,\log \,10 = 1.07\,volt\).
$A$. $\mathrm{Fe}$ $B$. $\mathrm{Mn}$ $C$. $\mathrm{Ni}$ $D$. $\mathrm{Cr}$ $E$. $\mathrm{Cd}$
Choose the correct answer from the options given below:
$Zn^{2+} + 2e^-$ $\longrightarrow$ $Zn (s) ; E^o = -0.76\,V$
$Ca^{2+} + 2e^-$ $\longrightarrow$ $Ca (s) ; E^o = -2.87\,V$
$Mg^{2+} + 2e^-$ $\longrightarrow$ $Mg (s) ; E^o = -2.36\,V$
$Ni^{2+} + 2e^-$ $\longrightarrow$ $Ni (s) ; E^o = -0.25\,V$
ધાતુઓની રિડક્શનકર્તા ઉર્જાનો ચઢતો ક્રમ કયો થશે?