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 ] = 0.855 M and [Ni2 ] = 0.0200 M.

Mg2+(aq) + 2e– ? Mg(s) –2.38 Calculate the cell potential for the following reaction as written at 25.00 , given that [Mg2+] = 0.855 M and [N21 0.0200 M. Standard reduction potentials can be found here. Mg(s)+Ni2+(aq) Mg2 + (aq)+Ni(s) Number Previous Give Up & View Solution O Check Answer Next Hint Use the Nernst equation RTinl where E is the cell potential, E is the standard cell potential, R is the gas contant (8.3145 J/(mol K)). Tis the Kelvin temperature, n is the number of electrons transferred, F is the Faraday constant (96485 J/(V mol)), and Q is the reaction quotient (product concentrations over reactant concentrations)

Explanation / Answer

Eo = Ecathode - Eanode = -0.26 - (-2.38) = 2.12 V

Q = [Mg2+]/[Ni2+]

Feed values given above,

Cell potential E,

E = 2.12 - (8.314 x 298/2 x 96485) ln(0.855/0.0200) = 2.072 V

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