A voltaic cell consists of a Hg/Hg2²⁺ electrode (E° = 0.85 V) and a Sn/Sn²⁺ electrode (E° = -0.14 V). Calculate [Sn²⁺] if [Hg₂²⁺] = 0.24 M and E_cell = 1.04 V at 25°C.

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The cell reaction is: Hg2²⁺ (s) + Sn²⁺ (aq) → Hg (l) + Sn²⁺ (s) The cell potential is given by the equation: E_cell = E°Hg/Hg2²⁺ - E°Sn/Sn²⁺ E_cell = 0.85 V - (-0.14 V) E_cell = 0.99 V We can use the Nernst equation to calculate the concentration of Sn²⁺: E_cell = E°cell - (RT/nF) ln Q where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the half-reactions, and F is Faraday's constant. At standard conditions (1 M concentrations and 25°C), the reaction quotient is equal to 1 and ln Q is equal to 0. Therefore: E_cell = E°cell 0.99 V = 0.85 V - (RT/nF) ln [Sn²⁺]/[Hg₂²⁺] Solving for [Sn²⁺], we get: [Sn²⁺] = [Hg₂²⁺] e^((E°cell - E_cell)/(RT/nF)) [Sn²⁺] = 0.24 M e^((0.85 V - 0.99 V)/(0.0083145 kJ/mol*K/2)(298.15 K)/(2*96485 C/mol)) [Sn²⁺] = 0.005 M Therefore, the answer is C, 0.005 M.

A Voltaic Cell Consists of a Hg/Hg22+ Electrode (E° =

A Voltaic Cell Consists of a Hg/Hg2²⁺ Electrode (E° =