Q.In the electrochemical cell $\text{Zn} \mid \text{ZnSO}_4\text{(0.01M)} \ \| \ \text{CuSO}_4\text{(1.0M)} \mid \text{Cu}$, the emf of this Daniel cell is $E_1$. When the concentration of $\text{ZnSO}_4$ is changed to 1.0M and that of $\text{CuSO}_4$ is changed to 0.01M, the emf changes to $E_2$. From the above, which one is the relationship between $E_1$ and $E_2$?

(a) A galvanic cell's Gibbs free energy change is directly tied to its EMF and, through the equilibrium constant, connects electrochemistry to chemical thermodynamics; the salt bridge is what makes the whole cell function by completing the circuit and keeping both half-cells electrically neutral. OR (b) diethyl ether reacts three different ways depending on the reagent — free-radical substitution with light, acid hydrolysis back to two alcohol molecules, and acid-catalysed cleavage with an acid chloride.

(a)(i) Thermodynamics of cell reactions: the maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction, related to the cell EMF by: $\Delta G = -nFE_{cell}$ where $n$ is the number of moles of electrons transferred and $F$ is the Faraday constant. Under standard conditions: $\Delta G° = -nFE°_{cell}$ Since $\Delta G°$ is also related to the equilibrium constant $K$ of the cell reaction by $\Delta G° = -RT\ln K$, combining gives $E°_{cell} = \dfrac{RT}{nF}\ln K$ linking the standard cell potential directly to the reaction's equilibrium constant. Additionally, the temperature coefficient of the cell EMF is related to the entropy change of the reaction: $\left(\dfrac{\partial E}{\partial T}\right)_P = \dfrac{\Delta S}{nF}$ and the enthalpy change follows from $\Delta H = \Delta G + T\Delta S$.

(a)(ii) Why a salt bridge is used in Galvanic cells: (1) it completes the internal electrical circuit of the cell, allowing ions to flow between the two half-cells (external circuit alone only allows electron flow through the wire); (2) it maintains electrical neutrality in both half-cell solutions as the cell reaction proceeds — e.g. as $Zn$ is oxidised to $Zn^{2+}$ in the anode compartment (building up positive charge) and $Cu^{2+}$ is reduced to $Cu$ in the cathode compartment (leaving excess negative charge), the salt bridge supplies/removes counter-ions to prevent charge build-up, which would otherwise stop the reaction; (3) it minimises (or eliminates) the liquid junction potential that would otherwise arise from direct contact between the two different electrolyte solutions.

OR (b) Reactions of diethyl ether:

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(a) Combining the general thermodynamic relation $\Delta G=\Delta G°+RT\ln Q$ with $\Delta G=-nFE$ yields the Nernst equation, which relates a cell/electrode's actual potential to its standard potential and reactant/product concentrations. OR (b) three classic dehydration reactions convert simple alcohols/polyols into an alkene, a cyclic ether, and an unsaturated aldehyde respectively.

(a) Derivation of the Nernst equation: consider a general electrode reaction $M^{n+}(aq) + ne^- \rightarrow M(s)$ The Gibbs free energy change for this process is related to the reaction quotient $Q$ by $\Delta G = \Delta G° + RT\ln Q$ Since the activity of the pure solid metal is taken as 1, $Q = 1/[M^{n+}]$. Also, the free energy change is related to the electrode potential by $\Delta G = -nFE$ (and $\Delta G° = -nFE°$), where $n$ is the number of electrons transferred and $F$ is the Faraday constant. Substituting: $-nFE = -nFE° + RT\ln\dfrac{1}{[M^{n+}]}$ Dividing throughout by $-nF$: $E = E° - \dfrac{RT}{nF}\ln\dfrac{1}{[M^{n+}]} = E° + \dfrac{RT}{nF}\ln[M^{n+}]$ For a full cell reaction (with reaction quotient $Q$ for the overall cell reaction), this generalises to: $E_{cell} = E°_{cell} - \dfrac{RT}{nF}\ln Q$ Converting from natural log to $\log_{10}$ ($\ln x = 2.303\log x$) and substituting the values of $R$, $F$ at $T=298\ K$: $E_{cell} = E°_{cell} - \dfrac{0.0591}{n}\log Q$ This is the Nernst equation.

OR (b)(i) Ethyl alcohol $\rightarrow$ ethene: dehydration of ethanol with concentrated sulphuric acid at $443\ K$ (or by passing vapour over hot $Al_2O_3$ at $623\ K$): $C_2H_5OH \xrightarrow[443K]{conc.\ H_2SO_4} CH_2=CH_2 + H_2O$

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(a) The Nernst equation, $E = E^\circ - (RT/nF)\ln Q$, follows from the thermodynamic relations $\Delta G = \Delta G^\circ + RT\ln Q$ and $\Delta G = -nFE$; (b) a catalyst is chemically unchanged, needed in small amount, doesn't shift equilibrium, and is often reaction-specific.

(a) Derivation of the Nernst equation:

For a general electrochemical cell reaction occurring with the transfer of $n$ moles of electrons, the free energy change under non-standard conditions is related to the standard free energy change by:

$\Delta G = \Delta G^\circ + RT\ln Q \qquad (1)$

where $Q$ is the reaction quotient for the cell reaction.

The free energy change of a cell reaction is related to the cell EMF by:

$\Delta G = -nFE \qquad (2)$

and at standard state:

$\Delta G^\circ = -nFE^\circ \qquad (3)$

where $F$ is the Faraday constant (96500 $C\,mol^{-1}$), $E$ is the cell EMF under the given (non-standard) conditions, and $E^\circ$ is the standard cell EMF.

Substituting (2) and (3) into (1):

$-nFE = -nFE^\circ + RT\ln Q$

Dividing throughout by $-nF$:

$E = E^\circ - \frac{RT}{nF}\ln Q$

This is the Nernst equation. At $T = 298\,K$, converting $\ln$ to $\log_{10}$ (multiply by 2.303) and substituting $R = 8.314\,J\,K^{-1}mol^{-1}$, $F = 96500\,C\,mol^{-1}$:

$E = E^\circ - \frac{0.0591}{n}\log_{10} Q$

For a single electrode reaction $M^{n+} + ne^- \rightarrow M$, this specialises to:

$E_{M^{n+}/M} = E^\circ_{M^{n+}/M} - \frac{0.0591}{n}\log_{10}\frac{1}{[M^{n+}]}$

The Nernst equation shows how the electrode/cell potential varies with the concentrations (activities) of the species involved, and it becomes $E = E^\circ$ only when all species are at unit concentration/activity ($Q=1$).

(b) Characteristics of a catalyst:

1. A catalyst remains chemically and quantitatively unchanged at the end of the reaction (its mass and chemical composition are the same before and after), even though its physical form may sometimes change (e.g. a large crystal may become powdered).
2. Only a small quantity of catalyst is generally required to catalyse even a large amount of reactants.
3. A catalyst cannot initiate a reaction that would not otherwise occur on its own (i.e. it cannot make a thermodynamically non-feasible reaction happen); it only changes the rate of a reaction that is already feasible. …

With Zn as the anode (oxidation) and Ni as the cathode (reduction), the standard cell potential works out to +0.51 V, giving a standard free energy change of −98.43 kJ/mol via ΔG° = −nFE°.

Cell and half-reactions: In $Zn\,|\,Zn^{2+}\,\|\,Ni^{2+}\,|\,Ni$ (standard cell notation: anode on left, cathode on right):

• Anode (oxidation): $Zn \rightarrow Zn^{2+} + 2e^-$, standard reduction potential $E^\circ(Zn^{2+}/Zn) = -0.76\ V$
• Cathode (reduction): $Ni^{2+} + 2e^- \rightarrow Ni$, standard reduction potential $E^\circ(Ni^{2+}/Ni) = -0.25\ V$
• Overall: $Zn + Ni^{2+} \rightarrow Zn^{2+} + Ni$

Standard EMF of the cell:

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ(Ni^{2+}/Ni) - E^\circ(Zn^{2+}/Zn)$

$E^\circ_{cell} = (-0.25\ V) - (-0.76\ V) = -0.25 + 0.76 = +0.51\ V$

The positive value confirms the cell reaction is spontaneous as written (Zn reduces Ni$^{2+}$), consistent with Zn being more easily oxidised (more negative $E^\circ$) than Ni.

Standard free energy change:

Number of electrons transferred in the balanced overall reaction, $n=2$ (both half-reactions involve 2 electrons); Faraday constant $F = 96500\ C\,mol^{-1}$. …

Applying the Nernst equation to the given half-cell potential and concentration and solving for $E^\circ$ gives a standard electrode potential of about $+0.36$ V for $Cu^{2+}/Cu$.

Half-cell reaction (reduction):

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$, so $n = 2$ electrons.

Nernst equation for this half-cell (at 298 K, using $\dfrac{2.303RT}{F} = 0.0591$ V):

$E = E^\circ - \dfrac{0.0591}{n}\log\dfrac{1}{[Cu^{2+}]}$

Given data: $E = +0.301$ V, $[Cu^{2+}] = 0.01$ M $= 10^{-2}$ M, $n = 2$.

Substituting:

$0.301 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{1}{10^{-2}}$

$\log\dfrac{1}{10^{-2}} = \log(10^{2}) = 2$

$0.301 = E^\circ - \dfrac{0.0591}{2}\times 2$

$0.301 = E^\circ - 0.0591$

Solving for $E^\circ$:

$E^\circ = 0.301 + 0.0591 = 0.3601\ \text{V}$

Rounding to three significant figures, $E^\circ \approx 0.36$ V.

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