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

Step 1: Part (a)(i): Write the zinc half-cell reaction.
\[ Zn^{2+} + 2e^- \rightarrow Zn \] $n = 2$, $E^\circ = -0.76$ V, $[Zn^{2+}] = 0.01$ M $= 10^{-2}$ M.
Step 2: Write the Nernst equation at $25^\circ C$.
For the half-cell $Zn^{2+}/Zn$: \[ E = E^\circ + \frac{0.0591}{n}\log[Zn^{2+}] \]
Step 3: Substitute values and calculate.
\[ E = -0.76 + \frac{0.0591}{2}\log(10^{-2}) = -0.76 + 0.02955 \times (-2) = -0.76 - 0.0591 = -0.8191 \approx -0.82 \text{ V} \]
Step 4: Part (a)(ii): Anode reaction in dry cell.
In a dry cell (Leclanche cell), the anode is zinc. Oxidation occurs: \[ Zn \rightarrow Zn^{2+} + 2e^- \text{ (oxidation at anode)} \]
Step 5: Cathode reaction in dry cell.
The cathode is a carbon rod with $MnO_2$ and $NH_4Cl$ paste. Reduction occurs: \[ 2MnO_2 + 2NH_4^+ + 2e^- \rightarrow Mn_2O_3 + 2NH_3 + H_2O \text{ (reduction at cathode)} \]
Step 6: Overall cell reaction and final answers.
Overall: \[ Zn + 2MnO_2 + 2NH_4^+ \rightarrow Zn^{2+} + Mn_2O_3 + 2NH_3 + H_2O \] EMF of dry cell $\approx 1.5$ V. Electrode potential of zinc half-cell in $0.01$ M $ZnSO_4$: \[ \boxed{E = -0.82 \text{ V}} \]