Question

The $E _{\text {red }}^{\circ}$ of $Ag,\, Cu ,\, Co$ and $Zn$ are $0.799$ $0.337,-0.277$ and $-0.762\, V$ respectively, which of the following cells will have maximum cell emf?

• A. $Zn \left| Zn ^{2+}(1 M ) \| Cu ^{2+}(1 M )\right| Cu$
• B. $Zn \left| Zn ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag$
• C. $Cu \left| Cu ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag$
• D. $Zn \left| Zn ^{2+}(1 M ) \| Co ^{2+}(1 M )\right| Co$

Answer 1

Answer: (B)

$Zn \left| Zn ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag$

Answer 2

(i) $Zn \left| Zn ^{2+}(1 M ) \| Ag ^{+}( lM )\right| Ag$
$Zn \left| Zn ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag$

given $E _{ Zn ^{2+} / Zn }^{\circ}=-0.762$
$E _{ Ag ^{+} / Ag }^{\circ}=0.799$

From Nernst equation,

$E = E ^{\circ}-\frac{0.059 l }{ n } \log \left[\frac{\text { Products }}{\text { Reactants }}\right]$

Cell reaction for the given cell is

$Zn +2 Ag ^{+} \longrightarrow Zn ^{2+}+2 Ag$
$E =E ^{\circ}{ }_{ Ag }^{+} / Ag ^{-} E ^{\circ} zn ^{2+} / Zn -\frac{0.059 l }{2} \log \frac{ Zn ^{2+}}{\left( Ag ^{+}\right)^{2}}$
$E = 0.799-(-0.762)-\frac{0.0591}{2} \log \left(\frac{1}{ l ^{2}}\right)$
$E = 1.569\, V$

(ii) $Cu \left| Cu ^{2+}(1 M ) \| Ag ^{+}( IM )\right| Ag$

Given $E _{ Cu ^{2+} / Cu }^{\circ} =0.337$
$E _{ Ag ^{\circ}}{ Ag } =0.799$

Cell reaction

$Cu +2 Ag ^{+} \longrightarrow Cu ^{2+}+2 Ag$
$E=E^{\circ}_{Ag^{+}/Ag} -E^{\circ}_{Cu^{2+}/Cu} - \frac{0.0591}{2} log \frac{Cu^{2+}}{\left(Ag^{+}\right)^{2}}$
$E = 0.799-0.337-\frac{0.0591}{2} \log \left(\frac{ l }{ l ^{2}}\right)$
$E = 0.462\, V$

(iii) $Zn \left| Zn ^{2+}(1 M )\right|\left| Co ^{2+}(1 M )\right| Co$

Given, $E _{ Zn ^{2+} / Zn }^{\circ}=-0.762 ; E _{ Co ^{2+} / Co }^{\circ}=-0.277$

Cell reaction $: Zn + Co ^{2+} \longrightarrow Zn ^{2+}+ Co$
$E = E ^{\circ} Co ^{2+} / Co ^{- E ^{\circ} Zn ^{2+} / Zn }-\frac{0.059 l }{2} \log \left[\frac{ Zn ^{2+}}{ Co ^{2+}}\right]$
$E =-0.277-(-0.762)-\frac{0.0591}{2} \log \left[\frac{1}{1}\right]$
$E =0.485\, V$

(iv) $Zn \left| Zn ^{2+}(1 M ) \| Cu ^{2+}( lM )\right| Cu$

Given, $E _{ Zn ^{2+} / Zn }^{\circ}=-0.762$
$E ^{\circ} Cu ^{2+} / Cu ^{2}=0.337$

Cell reaction $Zn + Cu ^{2+} \longrightarrow Zn ^{2+}+ Cu$
$E = E ^{\circ}{ Cu ^{2+} / Cu }- E ^{\circ} Zn ^{2+} / Zn ^{-} \frac{0.059 l }{2} \log \left[\frac{ Zn ^{2+}}{ Cu ^{2+}}\right]$
$E =0.337-(-0.762)-\frac{0.0591}{2} \log \left(\frac{1}{ l }\right)$
$E =1.099$

Hence, cell B has maximum emf.

So, the correct answer is (B): $Zn \left| Zn ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag$