Question

Answer the following: \[ (a)(i)\;\text{Calculate the electrode potential of a half-cell for zinc electrode dipping in }0.01M\;ZnSO_4\text{ solution at }25^\circ C. \] \[ E^\circ_{Zn^{2+}/Zn}=-0.76V,\quad \log 10=1 \] \[ (a)(ii)\;\text{Write anode, cathode and overall reaction involved in dry cell.} \] \[ (a)(iii)\;\text{Equilibrium constant }(K_c)\text{ is related to }E^\circ_{\text{cell}},\text{ but not to }E_{\text{cell}}.\text{ Why?} \]

Hint:

At equilibrium \(E_{\text{cell}}=0\), but \(E^\circ_{\text{cell}}\) is connected to \(K_c\) through \(\Delta G^\circ\).

Answer 1

Concept:
For a metal ion half-cell: \[ M^{n+}+ne^-\rightarrow M \] Nernst equation at \(25^\circ C\) is: \[ E=E^\circ+\frac{0.0591}{n}\log[M^{n+}] \]

Step 1: Write the zinc half-cell reaction.
\[ Zn^{2+}+2e^-\rightarrow Zn \] Here: \[ n=2 \] \[ [Zn^{2+}]=0.01M=10^{-2}M \]

Step 2: Apply Nernst equation.
\[ E=E^\circ+\frac{0.0591}{2}\log[Zn^{2+}] \] \[ E=-0.76+\frac{0.0591}{2}\log(10^{-2}) \] \[ \log(10^{-2})=-2 \] \[ E=-0.76+\frac{0.0591}{2}(-2) \] \[ E=-0.76-0.0591 \] \[ E=-0.8191V \] Therefore: \[ E=-0.819V \]

Step 3: Dry cell reactions.
In dry cell, zinc container acts as anode and graphite rod surrounded by \(MnO_2\) acts as cathode. At anode: \[ Zn\rightarrow Zn^{2+}+2e^- \] At cathode: \[ MnO_2+NH_4^++e^- \rightarrow MnO(OH)+NH_3 \] Overall reaction: \[ Zn+2MnO_2+2NH_4^+ \rightarrow Zn^{2+}+2MnO(OH)+2NH_3 \]

Step 4: Relation of K_c

with E^_

cell

.
The standard Gibbs free energy change is related to standard cell potential: \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] Also: \[ \Delta G^\circ=-RT\ln K_c \] Combining: \[ nFE^\circ_{\text{cell}}=RT\ln K_c \] Therefore: \[ E^\circ_{\text{cell}}=\frac{RT}{nF}\ln K_c \] Thus, \(K_c\) is related to \(E^\circ_{\text{cell}}\). But \(E_{\text{cell}}\) depends on concentration and changes according to Nernst equation. At equilibrium: \[ E_{\text{cell}}=0 \] Therefore, \(K_c\) is related to \(E^\circ_{\text{cell}}\), not to \(E_{\text{cell}}\). Hence: \[ E_{Zn^{2+}/Zn}=-0.819V \] \[ Zn\rightarrow Zn^{2+}+2e^- \] \[ MnO_2+NH_4^++e^-\rightarrow MnO(OH)+NH_3 \] \[ Zn+2MnO_2+2NH_4^+ \rightarrow Zn^{2+}+2MnO(OH)+2NH_3 \]