3.19 Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd (ii) Fe 2+(aq) + Ag+ (aq) → Fe3+(aq) + Ag(s) Calculate the ?rG? and equilibrium constant of the reactions. 

<p><strong> (1)</strong> Known - E⁰_Cr³⁺_/Cr = - 0.74 V</p><p>E⁰ cd²⁺ = - 0.40 V</p><p>? _rG⁰ =? K =? </p><p>The galvanic cell of the given reaction is written as - Cr_(s)|Cr³⁺_(aq)| Cd²⁺_(aq)|Cd_(s)→ Reaction 1</p><p>Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰</p><p>= - 0.40 - (- 0.74)</p><p>∴ E⁰ = + 0.34 V</p><p>To calculate the standard Gibb’s free energy? _rG⁰, we use, </p><p>? _rG⁰ = - nE⁰F → Equation 1</p><p>wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential. Substituting n = 6 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1, </p><p>E⁰ = + 0.34 V in Equation 1, we get, l</p><p>? _rG⁰ = - 6×0.34V×96487 C mol^-1</p><p>= - 196833.48 CV mol^-1</p><p>= - 196833.48 J mol^-1</p><p>∴? _rG⁰ = - 196.83348 kJ mol^-1</p><p>To find out the equilibrium constant, K, we use the formula, </p><p>log K =n E⁰ / 0.0591</p><p>=6 X 0.3 V / 0.0591</p><p>log K = 34.5177</p><p>K = antilog 34.5177</p><p>∴ K = 3.294 × 10³⁴</p><p>The standard Gibb’s free energy? _rG⁰ is - 196.83348 kJ mol ^–¹ and equilibrium constant, K is 3.294 × 10³⁴</p><p><strong> (2)</strong> Known -</p><p>E⁰ Fe³⁺/ Fe²⁺ = 0.77V</p><p>E⁰ Ag⁺ / Ag = 0.80 V</p><p>? _rG⁰ =? K =? </p><p>The galvanic cell of the given reaction is written as - Fe²⁺_(aq)|Fe³⁺_(aq)| Ag ⁺_(aq)|Ag_(s)→ Reaction 1</p><p>Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰</p><p>= 0.80 - (0.77)</p><p>∴ E⁰ = + 0.03 V</p><p>To calculate the standard Gibb’s free energy? _rG⁰, we use, </p><p>? _rG⁰ = - nE⁰F → Equation 1</p><p>wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential.</p><p>Substituting n = 1 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1, E⁰ = + 0.03V in Equation 1, we get, </p><p>? _rG⁰ = - 1×0.03V×96487 C mol^-1</p><p>= - 2894.61 CV mol ^-¹</p><p>= - 2894.61 J mol^-1</p><p>∴? _rG⁰ = - 2.89461 kJ mol^-1</p><p>To find out the equilibrium constant, K, we use the formula, </p><p>log K =n E⁰ / 0.0591</p><p>=1 X 0.03 V / 0.0591</p><p>log K = 0.5076</p><p>K = Antilog 0.5076</p><p>∴ K = 3.218</p><p>The standard Gibb’s free energy? _rG⁰ is - 2.89461 kJ mol ^–¹ and equilibrium constant, K is 3.218</p>

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