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295. Calculate the potential (emf.) of the cell
Cd | Cd²⁺ (0.10 M) || H⁺(0.20 M) | Pt, H₂ (0.5 atm)
(Given : E° _{Cd2+ / Cd} = – 0.403 V, R = 8.314 K^–1 K^–1 mol^–1, F = 96500 C mol^–1)

Cell reaction will be

Cd + 2 H^{+} \to {Cd}^{2 +} + H_{2} (hence n = 2)

E °_{cell} = E °_{right} - E °_{left} E °_{cell} = 0 - (- 0.403 = + 0.403 V)

E_{cell} = E °_{cell} + \frac{2.303}{nF} \log \frac{{[H^{+}]}^{2} [Cd]}{{pH}_{2} [{Cd}^{2 +}]}

= 0.403 + \frac{8.314 \times 298 \times 2.303}{2 \times 96500} \log \frac{(0.2)^{2}}{(0.5) (0.1)} = 0.403 + 0.0295 \log 0.8 = 0.403 - 0.0028

E_{cell} = 0.400 V

Equilibrium Constant from Nernst Equation: If the Daniell cell is short circuited, then we note that the reacton.
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
takes place and as time passes, the concentration of Zn²⁺ keeps on increasing while the concentration of Cu²⁺ keeps decreasing. At the same time, voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu²⁺ and Zn²⁺ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as

E_{cell}^{°} = \frac{2.303 RT}{nF} \log \frac{[{Cu}^{2 +}]}{[{Zn}^{2 +}]} E_{cell}^{°} = \frac{2.303 RT}{2 F} \log \frac{[{Cu}^{2 +}]}{[{Zn}^{2 +}]}

But at equilibrium [Zn²⁺] / [Cu²⁺] = K
and the above equation can be written as

E_{cell}^{°} = \frac{0.0591}{2} logK

(E° = 1.1 V)
K = 2 x 10³⁷ at 298
In general E°_cell = 2.303 RT / nF x log K

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Short Answer Type

296. Calculate ΔG° for the reaction

{Cu}^{2 +} (aq) + Fe (s) \to {Fe}^{2 +} (aq) + Cu (s) (Given E °_{{Cu}^{2 +} / Cu} = 0.34 V, E °_{{Fe}^{3 +} / Fe} = - 0.44 V, F = 96500 e {mol}^{- 1})

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297. The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the reaction:

Zn (s) + {Cu}^{2 +} (aq) \to {Zn}^{2 +} (aq) + Cu (s)

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298.

Calculate the Δ_r G° and the equilibrium constant for the reaction
2Cr(s) + 3Cd²⁺ (aq) 2Cr³⁺ (aq) + 3Cd(s)
(Given E°_Cr3+_/Cr = 0.74 V, E°_{Cd2+/ Cd} = – 0.40 V)

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299.

Calculate the cell e.m.f. and ΔG for the cell reaction at 298 K for the cell
Zn(s) | Zn²⁺ (0.0004 M) || Cd²⁺(0.2 M) | Cd(s)
(Given E°_Zn²⁺/Zn = 0.763 V, E°_{cd 2+ / cd} = 0.403 Vat 298 K, F = 96500 C mol^–1)

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300. Calculate the cell emf and AG for the cell rection at 25°C for the cell
Zn(s) | Zn²⁺ (0.0004 M) || Cd²⁺ (0.2 M) | Cd(s) E° values at 25°C, Zn²⁺/Zn = – 0.763 V, Cd²⁺/ Cd = – 0.403 V F = 96500 C mol^–1, R = 8.314 J K^–1 mol^–1

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