Calculate the cell potential for the following reaction as written at 25.00 °C,

Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.771 M and [Fe2 ] = 0.0150 M. Standard reduction potentials can be found here

.Mg(s)+ Fe2+(aq) (double headed arrow) Mg2+(aq) +Fe(s)

Calculate the cell potential for the following reaction as written at 25.00 pc, given that IMg 0.771 M and [Fe 0.0150 M. Standard reduction potentials can be found here Number O Previous Give Up & View Solution check Answer Nex Exit Hint Use the Nemst equation RT where Eis the cell potentia E is the standard cell potential, Ris the gas contant (8-3145 Jl mol K), Tis the Kelvin temperature, the number of electrons transferred, Fis the Faraday constant (96485 JIV mol Q is the reaction quotient (product concentrations over reactant concentrations). Note that n is symbolized as ve orzin some textbooks.

Explanation / Answer

E0Mg2+/Mg = -2.37 V

E0Fe2+/Fe = -0.44 V

E0cell = E0cathode - E0anode

       = -0.44 - (-2.37) = 1.93 V

Ecell = E0 - 0.0591/nlog(Mg2+/Fe2+)

      = 1.93 - (0.0591/2)log(0.771/0.015)

      = 1.88 V

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