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Answer the following.

Answer the following: Calculate the values of and for the following cell reaction at : (Given: ; …

Chemistry20263 marksNumerical
Answer the following:
(a)[2.0]
Calculate the values of $E_{\text{cell}}$ and $\Delta G$ for the following cell reaction at $25^\circ\text{C}$: $\text{Zn(s)} / \text{Zn}^{2+}(0\cdot0004\text{ M}) \parallel \text{Cd}^{2+}(0\cdot2\text{ M}) / \text{Cd(s)}$ (Given: $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0\cdot763\text{ V}$; $E^\circ(\text{Cd}^{2+}/\text{Cd}) = -0\cdot403\text{ V}$, $1\text{ Faraday} = 96,500\text{ coulombs}$, $R = 8\cdot314\text{ J K}^{-1}\text{ mol}^{-1}$)
(b)[1.0]
Calculate how long it will take to deposit $1\cdot0\text{ g}$ of chromium when a current of $1\cdot25\text{ ampere}$ flows through a solution of chromium (III) sulphate. (Atomic weight of $\text{Cr} = 52$, $1\text{ Faraday} = 96,500\text{ coulombs}$.)

Answer

Answer (a)

Official answer key

Formulas used: Standard emf of a cell: E0cell = E0cathode - E0anode with E0anode = -0.763 V, E0cathode = -0.403 V, so E0cell = 0.36 V Nernst equation at temperature T: E = E0 - 2.303*R*T/(n*F)*log10(Q) with Q = 0.002, R = 8.314 J K^-1 mol^-1, T = 298 K, n = 2, E0 = 0.36 V, so E = 0.44 V Gibbs energy from emf: dG = -n*F*E with E = 0.44 V, F = 96500 C mol^-1, n = 2, so dG = -84920 J mol^-1

Cell reaction: $\mathrm{Zn + Cd^{2+} \rightarrow Zn^{2+} + Cd}$, $n = 2$ $E^\circ_{cell} = -0.403 - (-0.763) = 0.36\ \mathrm{V}$ $E_{cell} = 0.36 - \dfrac{0.059}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cd^{2+}}]} = 0.36 - \dfrac{0.059}{2}\log\dfrac{0.0004}{0.2} = 0.36 + 0.08 = 0.44\ \mathrm{V}$ $\Delta G = -nFE = -2\times96500\times0.44 = -84920\ \mathrm{J} = -84.92\ \mathrm{kJ}$

Final answer: 0.44 V

Answer (b)

Official answer key

Formula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, I = 1.25 A, M = 52 g mol^-1, n = 3, w = 1.0 g, so t = 4453.8 s

$\mathrm{Cr^{3+} + 3e^- \rightarrow Cr}$: 52 g of Cr needs $3\times96500$ C, so 1.0 g needs $Q = \dfrac{3\times96500}{52} = 5567.3\ \mathrm{C}$ $t = \dfrac{Q}{I} = \dfrac{5567.3}{1.25} = 4453.8\ \mathrm{s}$

Final answer: 4453.8 s

Electrochemistry

From ISC 2026 Specimen Chemistry Paper 1, question 16(ii).

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