Electrochemistry Set-1

Q1. A galvanic cell at $298\,\mathrm{K}$ is constructed as $\mathrm{Zn}(s)\;|\;\mathrm{Zn}^{2+}(0.010\,\mathrm{M})\;||\;\mathrm{Cu}^{2+}(0.100\,\mathrm{M})\;|\;\mathrm{Cu}(s)$. 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 $E$ at $298\,\mathrm{K}$ using the Nernst equation (use $0.0591$ for $RT/F$).

Q2. A concentration cell is formed by two hydrogen electrodes at $298\,\mathrm{K}$: $\mathrm{Pt}|\mathrm{H}_2(1\,\mathrm{atm})|\mathrm{H}^+(10^{-3}\,\mathrm{M})\;||\;\mathrm{H}^+(10^{-6}\,\mathrm{M})|\mathrm{H}_2(1\,\mathrm{atm})|\mathrm{Pt}$. Using the Nernst equation, compute the emf $E$ (magnitude) and identify which electrode (the $10^{-3}\,\mathrm{M}$ side or the $10^{-6}\,\mathrm{M}$ side) will act as the cathode.

Q3. An electrolytic deposition is attempted from an aqueous solution containing $[\mathrm{Ag}^+]=0.010\,\mathrm{M}$, $[\mathrm{Cu}^{2+}]=0.10\,\mathrm{M}$ and $[\mathrm{Na}^+]=1.0\,\mathrm{M}$. Using the Nernst equation at $298\,\mathrm{K}$ and given $E^\circ_{\mathrm{Ag}^+/\mathrm{Ag}}=+0.80\,\mathrm{V}$, $E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}=+0.34\,\mathrm{V}$ and $E^\circ_{\mathrm{Na}^+/\mathrm{Na}}=-2.71\,\mathrm{V}$, which species will be reduced (deposit) first on the cathode?

Q4. Two one-electron half-cells have standard reduction potentials $E^\circ_X=-0.10\,\mathrm{V}$ for $X^+ + e^- \rightleftharpoons X$ and $E^\circ_Y=+0.20\,\mathrm{V}$ for $Y^+ + e^- \rightleftharpoons Y$. A galvanic cell assembled as $X|X^+(1.0\,\mathrm{M})||Y^+(10^{-6}\,\mathrm{M})|Y$ at $298\,\mathrm{K}$ exhibits an experimental emf of approximately $-0.055\,\mathrm{V}$ (negative). Which of the following best explains this observation?

Q5. Assertion (A): For a reversible concentration cell of the type $M|M^{z+}(c_1)||M^{z+}(c_2)|M$, the emf is given by $E=\dfrac{0.0591}{z}\log\!\left(\dfrac{c_2}{c_1}\right)$ and is independent of the electrode surface area. Reason (R): The Nernst equation arises from equality of chemical potentials and depends only on activities (concentrations), not on geometric factors such as surface area; electrode area affects kinetics (current) but not the equilibrium emf.

Q6. Calculate the emf at 298 K of the cell $Zn|Zn^{2+}(0.010\,\mathrm{M})||Cu^{2+}(0.0010\,\mathrm{M})|Cu$. Given $E^\circ_{Cu^{2+}/Cu}=+0.34\ \mathrm{V}$ and $E^\circ_{Zn^{2+}/Zn}=-0.76\ \mathrm{V}$.

Q7. A Daniell cell at 298 K is $Zn|Zn^{2+}(0.010\,\mathrm{M})||Cu^{2+}(0.200\,\mathrm{M})|Cu$. The cathode compartment contains $250\ \mathrm{mL}$ of the Cu solution. By controlled electrolysis $0.643\ \mathrm{g}$ of Cu is removed from the cathode compartment (assume negligible volume change). Using $E^\circ_{Cu^{2+}/Cu}=+0.34\ \mathrm{V}$ and $E^\circ_{Zn^{2+}/Zn}=-0.76\ \mathrm{V}$, the new emf of the cell at 298 K is closest to:

Q8. Consider the cell $Pt|H_{2}(1\,\mathrm{atm})|H^{+}(aq)||Cu^{2+}(0.010\,\mathrm{M})|Cu$ at 298 K. The measured emf is $E_{\text{cell}}=0.320\ \mathrm{V}$. Given $E^\circ_{Cu^{2+}/Cu}=+0.34\ \mathrm{V}$, calculate the pH of the hydrogen electrode compartment.

Q9. Assertion (A): During electrolysis of concentrated aqueous NaCl on an inert anode, $Cl_{2}$ is evolved at the anode rather than $O_{2}$, even though standard potentials under some conditions suggest oxygen evolution could be thermodynamically competitive.
Reason (R): Large overpotential for oxygen evolution on common anode materials (and the high Cl⁻ concentration shifting the chloride oxidation potential by the Nernst equation) makes chloride oxidation kinetically and electrochemically favored.

Q10. A cell is made with $Fe^{3+}/Fe^{2+}$ and $Cu^{2+}/Cu$ half-cells at 298 K. Initial concentrations are $[Fe^{3+}]=0.010\ \mathrm{M}$, $[Fe^{2+}]=0.0010\ \mathrm{M}$ and $[Cu^{2+}]=0.010\ \mathrm{M}$. Metal electrodes are $Fe(s)$ and $Cu(s)$. Given $E^\circ_{Fe^{3+}/Fe^{2+}}=+0.77\ \mathrm{V}$ and $E^\circ_{Cu^{2+}/Cu}=+0.34\ \mathrm{V}$, when the cell reaches equilibrium ($E_{\text{cell}}=0$) the ratio $\dfrac{[Fe^{3+}]}{[Fe^{2+}]}$ will be approximately: