Electrochemistry Set-5

Q1. A galvanic cell at 25°C is made of the half-cells Zn(s)|Zn^{2+}(0.010 M) and Cu^{2+}(1.0 M)|Cu(s). Given $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=+0.34\ \mathrm{V}$ and $E^\circ_{\mathrm{Zn}^{2+}/\mathrm{Zn}}=-0.76\ \mathrm{V}$, calculate the cell emf using the Nernst equation $E=E^\circ-\dfrac{0.0591}{n}\log Q$ (take $n=2$).

Q2. At 25°C, a galvanic cell consists of a standard hydrogen electrode (Pt|H_2(1 atm)|H^+(aq)) and a silver electrode (Ag^+(0.0010 M)|Ag). The measured cell emf (with Ag electrode as cathode) is $0.455\ \mathrm{V}$. Given $E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}}=+0.80\ \mathrm{V}$, calculate the pH of the hydrogen electrode solution (use the Nernst equation for the overall reaction).

Q3. At 25°C, solutions have $[\mathrm{Fe}^{2+}]=1.0\ \mathrm{M}$, $[\mathrm{Fe}^{3+}]=0.10\ \mathrm{M}$ and $[\mathrm{Cu}^{2+}]=0.10\ \mathrm{M}$. Given $E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}=+0.77\ \mathrm{V}$ and $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=+0.34\ \mathrm{V}$, will the reaction $\mathrm{Cu}^{2+} + 2\mathrm{Fe}^{2+} \rightarrow \mathrm{Cu}(s) + 2\mathrm{Fe}^{3+}$ proceed spontaneously? Calculate the cell emf to justify your answer (use appropriate Nernst expressions for each half-cell).

Q4. Assertion: For the electrode system Ag(s)|AgCl(s)|Cl^-(aq) at 25°C, the electrode potential is independent of the mass of solid AgCl present provided some AgCl(s) remains undissolved.
Reason: The half‑reaction $\mathrm{AgCl}(s) + e^- \rightarrow \mathrm{Ag}(s) + \mathrm{Cl}^-(aq)$ has the Nernst form $E=E^\circ_{\mathrm{AgCl/Ag}}-0.0591\log[\mathrm{Cl}^-]$, so the amount of solid does not appear in the expression.

Q5. At 25°C a concentration cell is made with two copper electrodes: left half‑cell Cu(s)|Cu^{2+} with $[\mathrm{Cu}^{2+}]_{\text{left}}=1.0\times10^{-6}\ \mathrm{M}$; right half‑cell Cu(s)|Cu^{2+} contains total copper $C_T=1.0\times10^{-3}\ \mathrm{M}$ and $[\mathrm{NH}_3]=0.10\ \mathrm{M}$, where copper forms $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ via $\mathrm{Cu}^{2+}+4\mathrm{NH}_3\rightleftharpoons[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$ with $K_f=1.0\times10^{12}$. Assuming activities ≈ concentrations and $n=2$, calculate the emf of the cell (left electrode initially has higher free $[\mathrm{Cu}^{2+}]$).