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Electrochemistry

Short Answer Type

291. Determine the equilibrium constant of the reaction at 298 K.

2 {Fe}^{3 +} + {Sn}^{2 +} ⇋ 2 {Fe}^{2 +} + {Sn}^{4 +}

From the obtained value of the equilibrium constant, predict whether Sn²⁺ ions can reduce Fe³⁺to Fe²⁺ quantitatively or not.

E °_{{Sn}^{4 +} / {Sn}^{2 +}} = 0.15 V and E °_{{Fe}^{3 +} / {Fe}^{2 +}} = + 0.77 V

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292. Calculate the equilibrium constant for the reaction at 298 K

Cu (s) + {Cl}_{2} (g) ⇌ {CuCl}_{2} (aq)

Given:

E °_{{Cu}^{2 +} / Cu} = 0.34 V; E °_{{Cl}^{2 +} / {Cl}^{-}} = 1.36 V, R = 8.314 J K^{- 1} {mol}^{- 1}

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293. Calculate the equilibrium constant K_c for the reaction at 298 K.

3 {Sn}^{4 +} + 2 Cr ⇋ 3 {Sn}^{2 +} + 2 {Cr}^{3 +}

(Given E °_{{Sn}^{4 +} / {Sn}^{2 +}} = 0.15 V; E °_{{Cr}^{3 +} / Cr} = - 074 V) . = 0.34 V;

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294. Calculate the equilibrium constant for the cell reaction:

4 {Br}^{-} + O_{2} + 4 H^{+} \to 2 {Br}_{2} + 2 H_{2} O

(Given

E °_{cell} = 0.16

)

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Long Answer Type

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)

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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)

(i) According to nernst equation:
   $E = E ° - \frac{0.0591}{n} \log \frac{[{Zn}^{2 +} (aq)]}{[{Cd}^{2 +} (aq)]} E °_{cell} = E °_{{Cd}^{2 +} / Cd} - E °_{{Zn}^{2 +} / Zn} = (0.403) - (- 0.763) = 0.36 V$
   $[{Zn}^{+} (aq)] = 0.0004 M, [{Cd}^{2 +} (aq)] = 0.2 M, n = 2$
   $E = (0.36) - \frac{(0.0591 V)}{2} \log \frac{0.0004}{0.2} = 0.36 - \frac{(0.0591 V)}{2} \times (- 2.69990) = 0.36 V + 0.08 = 0.44 V$
(ii)
   $∆ G = - {nFE}_{cell} E °_{cell} = 0.44 V, n = 2 mol, F = 96500 c {mol}^{- 1} ∆ G = - 2 (mol) \times (96500 C {mol}^{- 1}) \times (0.44 V) = - 84920 CV = - 84920 J .$

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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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