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1 (a) What is electrochemistry?

Chemistry

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1 (a) What is electrochemistry?
Electrochemistry is the branch of chemistry that studies the relationship between electrical energy and chemical changes. It deals with chemical reactions that produce electricity (voltaic or galvanic cells) and chemical reactions that are driven by electricity (electrolytic cells).

1 (b) Derive the Nernst equation and explain all terms.

Step 1: The change in Gibbs free energy (ΔG\Delta G) for a reaction is related to the standard Gibbs free energy change (ΔG∘\Delta G^{\circ}) and the reaction quotient (QQ) by the equation:

ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^{\circ} + RT \ln Q

Step 2: For an electrochemical cell, the Gibbs free energy change is also related to the cell potential (EcellE_{cell}) and the standard cell potential (Ecell∘E^{\circ}_{cell}) by:

ΔG=−nFEcell\Delta G = -nFE_{cell} ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{cell}

where nn is the number of moles of electrons transferred, and FF is Faraday's constant.

Step 3: Substitute these expressions for ΔG\Delta G and ΔG∘\Delta G^{\circ} into the first equation:

−nFEcell=−nFEcell∘+RTln⁡Q-nFE_{cell} = -nFE^{\circ}_{cell} + RT \ln Q

Step 4: Divide the entire equation by −nF-nF:

Ecell=Ecell∘−RTnFln⁡QE_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln Q

This is the Nernst equation. For the given reversible reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the reaction quotient QQ is:

Q=[C]c[D]d[A]a[B]bQ = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Substituting QQ into the Nernst equation:

Ecell=Ecell∘−RTnFln⁡([C]c[D]d[A]a[B]b)E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln \left( \frac{[C]^c [D]^d}{[A]^a [B]^b} \right)

Step 5: Convert the natural logarithm (ln⁡\ln) to base-10 logarithm (log⁡\log) using ln⁡x=2.303log⁡x\ln x = 2.303 \log x:

Ecell=Ecell∘−2.303RTnFlog⁡([C]c[D]d[A]a[B]b)E_{cell} = E^{\circ}_{cell} - \frac{2.303 RT}{nF} \log \left( \frac{[C]^c [D]^d}{[A]^a [B]^b} \right)

Explanation of terms:

  • EcellE_{cell}: The cell potential (or electromotive force, EMF) under non-standard conditions, measured in Volts (V).
  • Ecell∘E^{\circ}_{cell}: The standard cell potential under standard conditions (1 M concentration for solutions, 1 atm pressure for gases, 298 K temperature), measured in Volts (V).
  • RR: The ideal gas constant, 8.314 J mol−1 K−18.314 \text{ J mol}^{-1} \text{ K}^{-1}.
  • TT: The absolute temperature in Kelvin (K).
  • nn: The number of moles of electrons transferred in the balanced redox reaction.
  • FF: Faraday's constant, the charge of one mole of electrons, approximately 96485 C mol−196485 \text{ C mol}^{-1} (given as 96500 C mol−196500 \text{ C mol}^{-1} in constants).
  • QQ: The reaction quotient, which expresses the relative amounts of products and reactants at any given time. For the reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, Q=[C]c[D]d[A]a[B]bQ = \frac{[C]^c [D]^d}{[A]^a [B]^b}.

1 (c) Calculate the potential of the zinc electrode.

Step 1: Identify the half-reaction and given values.
The half-reaction for zinc reduction is:

Zn2+(aq)+2e−⟶Zn(s)Zn^{2+}(aq) + 2e^- \longrightarrow Zn(s)

Given:

  • Standard electrode potential, E∘(Zn2+/Zn)=0.52 VE^{\circ}(Zn^{2+}/Zn) = 0.52 \text{ V}
  • Temperature, T=27 °C=27+273.15=300.15 KT = 27 \text{ °C} = 27 + 273.15 = 300.15 \text{ K}
  • Initial concentration of zinc sulphate, [ZnSO4]=0.1 M[ZnSO_4] = 0.1 \text{ M}
  • Dissociation of zinc sulphate = 76%76 \%
  • Number of electrons transferred, n=2n = 2
  • Faraday's constant, F=96500 C mol−1F = 96500 \text{ C mol}^{-1}
  • Gas constant, R=8.314 J mol−1 K−1R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}

Step 2: Calculate the actual concentration of Zn2+Zn^{2+} ions.
Since ZnSO4ZnSO_4 dissociates into Zn2+Zn^{2+} and SO42−SO_4^{2-}, and it's 76%76\% dissociated:

[Zn2+]=[ZnSO4]×dissociation percentage[Zn^{2+}] = [ZnSO_4] \times \text{dissociation percentage} [Zn2+]=0.1 M×0.76=0.076 M[Zn^{2+}] = 0.1 \text{ M} \times 0.76 = 0.076 \text{ M}

Step 3: Apply the Nernst equation for a half-cell.
For a reduction half-reaction Mn++ne−⟶M(s)M^{n+} + ne^- \longrightarrow M(s), the Nernst equation is:

E=E∘−RTnFln⁡(1[Mn+])E = E^{\circ} - \frac{RT}{nF} \ln \left( \frac{1}{[M^{n+}]} \right)

Or, using log⁡10\log_{10} and substituting 2.303×R×T/F2.303 \times R \times T / F at 298 K298 \text{ K} as 0.0592 V0.0592 \text{ V}:

E=E∘−0.0592nlog⁡10(1[Mn+])(at 298 K)E = E^{\circ} - \frac{0.0592}{n} \log_{10} \left( \frac{1}{[M^{n+}]} \right) \quad \text{(at 298 K)}

Since the temperature is 300.15 K300.15 \text{ K}, we must use the full RT/nFRT/nF term.

E=E∘−RTnFln⁡(1[Zn2+])E = E^{\circ} - \frac{RT}{nF} \ln \left( \frac{1}{[Zn^{2+}]} \right) E=0.52 V−(8.314 J mol−1 K−1)(300.15 K)(2)(96500 C mol−1)ln⁡(10.076)E = 0.52 \text{ V} - \frac{(8.314 \text{ J mol}^{-1} \text{ K}^{-1})(300.15 \text{ K})}{(2)(96500 \text{ C mol}^{-1})} \ln \left( \frac{1}{0.076} \right) E=0.52 V−2495.0571 J mol−1193000 C mol−1ln⁡(13.1579)E = 0.52 \text{ V} - \frac{2495.0571 \text{ J mol}^{-1}}{193000 \text{ C mol}^{-1}} \ln (13.1579) E=0.52 V−(0.012927 V)×(2.577)E = 0.52 \text{ V} - (0.012927 \text{ V}) \times (2.577) E=0.52 V−0.0333 VE = 0.52 \text{ V} - 0.0333 \text{ V} E=0.4867 VE = 0.4867 \text{ V}

The potential of the electrode is 0.487 V\boxed{\text{0.487 V}}.

2 (a) Use an unambiguous diagram to illustrate how a beam of radiation is attenuated by an absorbing substance, and define the terms PtP_t and PoP_o.

Step 1: Diagram.

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