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Solve the following.
Calculate the value of Δ G° for the following cell at 298 K: 2Cr(s) + 3Cd²⁺(1M) → 2Cr³⁺(1M) +…
Calculate the value of $\Delta G^\circ$ for the following cell at $298\text{ K}$:
$2\text{Cr(s)} + 3\text{Cd}^{2+}(1\text{M}) \rightarrow 2\text{Cr}^{3+}(1\text{M}) + 3\text{Cd(s)}$
Given $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0\cdot74\text{ V}$, $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0\cdot40\text{ V}$
$1\text{ faraday} = 96,500\text{ coulombs}$
Answer
Answer
Formulas used: Standard emf of a cell: E0cell = E0cathode - E0anode with E0anode = -0.74 V, E0cathode = -0.40 V, so E0cell = 0.34 V Gibbs energy from emf: dG = -n*F*E with E = 0.34 V, F = 96500 C mol^-1, n = 6.0, so dG = -196860 J mol^-1
Cathode (reduction): $\mathrm{Cd^{2+}/Cd}$; anode (oxidation): $\mathrm{Cr/Cr^{3+}}$.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.40 - (-0.74) = 0.34\ \mathrm{V}$
$n = 6$ (2 Cr lose 6 electrons)
$\Delta G^\circ = -nFE^\circ_{cell} = -6 \times 96500 \times 0.34 = -196860\ \mathrm{J\ mol^{-1}}$
Final answer: -196.86 kJ mol^-1