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Exercises · 2.4

Q.Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

(i) 2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd
(ii) Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)
Calculate the ΔrG∘\Delta_rG^\circ and equilibrium constant of the reactions.
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The standard cell potential is the difference between the standard reduction potentials of the cathode and anode. For each reaction, we identify the half-reactions, look up their standard potentials, and then use Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}. From Ecell∘E^\circ_{\text{cell}}, we calculate ΔrG∘=−nFEcell∘\Delta_r G^\circ = -nFE^\circ_{\text{cell}} and the equilibrium constant KK using ln⁡K=nFEcell∘RT\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}.

The Core Idea: Why Cell Potential Tells Us About Spontaneity

A galvanic cell works because electrons flow spontaneously from a stronger reducing agent (anode, where oxidation happens) to a weaker one (cathode, where reduction happens). The driving force is the difference in their tendencies to gain electrons — measured as the standard reduction potential, E∘E^\circ.

The Nernst equation at standard conditions gives us the cell potential directly:

Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

A positive Ecell∘E^\circ_{\text{cell}} means the reaction is spontaneous. From there, the Gibbs free energy change tells us the maximum useful work obtainable:

ΔrG∘=−nFEcell∘\Delta_r G^\circ = -nFE^\circ_{\text{cell}}

And the equilibrium constant KK tells us how far the reaction goes:

ΔrG∘=−RTln⁡K⇒ln⁡K=nFEcell∘RT\Delta_r G^\circ = -RT \ln K \quad \Rightarrow \quad \ln K = \frac{nFE^\circ_{\text{cell}}}{RT}

At 298 K, using F=96485 C mol−1F = 96485\ \text{C mol}^{-1} and R=8.314 J mol−1K−1R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}, we often use the convenient form:

log⁡10K=nEcell∘0.059\log_{10} K = \frac{nE^\circ_{\text{cell}}}{0.059}

Let's apply this to each reaction.

Reaction (i): 2Cr(s)+3Cd2+(aq)→2Cr3+(aq)+3Cd(s)2Cr(s) + 3Cd^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Cd(s)

1. Identify the half-reactions

  • Oxidation (anode): Cr metal loses electrons to become Cr³⁺.

    Cr(s)→Cr3+(aq)+3e−Cr(s) \rightarrow Cr^{3+}(aq) + 3e^-

    Standard reduction potential (for the reverse): ECr3+/Cr∘=−0.74 VE^\circ_{Cr^{3+}/Cr} = -0.74\ \text{V}

  • Reduction (cathode): Cd²⁺ gains electrons to become Cd metal.

    Cd2+(aq)+2e−→Cd(s)Cd^{2+}(aq) + 2e^- \rightarrow Cd(s)

    Standard reduction potential: ECd2+/Cd∘=−0.40 VE^\circ_{Cd^{2+}/Cd} = -0.40\ \text{V}

Watch out

A common mistake is to use the oxidation potential directly. Always use reduction potentials from the table, and subtract the anode's reduction potential from the cathode's.

2. Calculate Ecell∘E^\circ_{\text{cell}}

Ecell∘=Ecathode∘−Eanode∘=(−0.40)−(−0.74)=+0.34 VE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.40) - (-0.74) = +0.34\ \text{V}

The positive value confirms the reaction is spontaneous as written.

3. Determine nn, the number of electrons transferred

Look at the balanced equation: 2 Cr atoms each lose 3 electrons → total 6 electrons lost. 3 Cd²⁺ ions each gain 2 electrons → total 6 electrons gained. So n=6n = 6.

4. Calculate ΔrG∘\Delta_r G^\circ

ΔrG∘=−nFEcell∘=−6×96485×0.34\Delta_r G^\circ = -nFE^\circ_{\text{cell}} = -6 \times 96485 \times 0.34

ΔrG∘=−196,829.4 J mol−1≈−196.8 kJ mol−1\Delta_r G^\circ = -196,829.4\ \text{J mol}^{-1} \approx -196.8\ \text{kJ mol}^{-1}

The negative sign means the reaction is spontaneous and can do useful work.

5. Calculate the equilibrium constant KK

Using ln⁡K=nFEcell∘RT\ln K = \frac{nFE^\circ_{\text{cell}}}{RT}:

ln⁡K=6×96485×0.348.314×298\ln K = \frac{6 \times 96485 \times 0.34}{8.314 \times 298}

ln⁡K=196829.42477.572≈79.45\ln K = \frac{196829.4}{2477.572} \approx 79.45

So K=e79.45K = e^{79.45}. That's an astronomically large number — the reaction goes essentially to completion.

Using the base-10 shortcut at 298 K:

log⁡10K=nEcell∘0.059=6×0.340.059=2.040.059≈34.58\log_{10} K = \frac{nE^\circ_{\text{cell}}}{0.059} = \frac{6 \times 0.34}{0.059} = \frac{2.04}{0.059} \approx 34.58

So K≈1034.58K \approx 10^{34.58}, consistent with the above.

Tip

When Ecell∘E^\circ_{\text{cell}} is positive and nn is large, KK becomes enormous — the reaction is product-favoured overwhelmingly.

Reaction (ii): Fe2+(aq)+Ag+(aq)→Fe3+(aq)+Ag(s)Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)

1. Identify the half-reactions

  • Oxidation (anode): Fe²⁺ loses an electron to become Fe³⁺. …
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