An electrochemical cell is constructed such that on one side a pure nickel electrode is in contact with a solution containing Ni2 + ions at a concentration of 2 * 10-3 M. The other cell half consists of a pure Fe electrode that is immersed in a solution of Fe2 + ions having a concentration of 0.1 M. At what temperature will the potential between the two electrodes be + 0.140 V?

676°C

Explanation:

The value of the potential between the electrodes can be calculated by the Nernst equation:

E = E° - (RT/nF) * lnQ

Where E° is the standard reduction potential of the cell, R is the ideal gas constant (8.315 J/mol. K), T is the temperature, n is the number of electrons being replaced in the reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction coefficient ([cathode]/[anode]).

The reaction will happen with oxidation (loss of electrons) at the anode and a reduction (gain of electrons) in the cathode. It will go from the most concentrated solution to the less concentrated (anode to cathode). The value of E° = Ecathode - Eanode, and the half-reactions are:

Ni (s) → Ni⁺² (aq) + 2e⁻ E = - 0.24 V

Fe (s) → Fe⁺² (aq) + 2e⁻ E = - 0.44 V

So, the solution of Fe is the cathode, and the solution of Ni is the anode.

E = - 0.44 - (-0.24) = - 0.20 V

n = 2

0.140 = - 0.20 - (8.315T/2*96485) * ln (2x10⁻³/0.1)

0.160 = 1.686x10⁻⁴T

T = 949.17 K

T = 676°C