Calculate the cell potential for the following reaction as written at 25.00 degr

Question

Calculate the cell potential for the following reaction as written at 25.00 degree C, given that [Zn^2+] = 0.848 M and [Ni^2+] = 0.0170 M. Standard reduction potentials can be found here. Zn(s) + Ni^2+ (aq) Zn^2+ (aq)+ Ni(s) First, calculate the standard cell potential, E degree. Then use the Nemst equation E = E degree - RT/nF ln (Q) where n is the number of electrons transferred, F is the Faraday constant (F = 96485 J/(V middot mol)), R is the gas constant (R = 8.3145 J/(mol middot K)), T is the absolute temperature, and Q is the reaction quotient (product concentrations over reactant concentrations). Note that n is sometimes symbolized as v_e or z in some textbooks.

Explanation / Answer

E0cell = Ec - Ea

    = (-0.25-(-0.767) = 0.517 v

E = E0 - (0.0591/n)log(Zn^2+ / Ni^2+)

= 0.517 - (0.0591/2)log(0.848/0.017)

= 0.467 v

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