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

The Correct Option is B

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