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An electrochemical cell is composed of pure copper and pure cadmium electrodes immersed in solutions of their respective divalent ions. For a 6.5 × 10-2 M concentration of Cd2+, the cadmium electrode is oxidized, yielding a cell potential of 0.775 V. Calculate the concentration of Cu2+ ions if the temperature is 25°C.

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From the calcuation, the concentration of the copper II ion is 0.98 M.

What is an electrochemical cell?

An electrochemical cell is one in which current is produced by means of a chemical reaction. In this cell, the cell reaction is;

Cd(s) + Cu^2+(aq) ----> Cd^2+(aq) + Cu(s)

E°cell = 0.34 V - (-0.40V) = 0.74 V

Using the Nernst equation;

0.775= 0.74 - 0.0592/2 log(6.5 × 10-2)/[Cu^2+]

0.775 - 0.74 =  - 0.0592/2 log(6.5 × 10-2)/[Cu^2+]

0.035 = -0.0296 log(6.5 × 10-2)/[Cu^2+]

0.035 /-0.0296 = log(6.5 × 10-2)/[Cu^2+]

-1.18 =  log(6.5 × 10-2)/[Cu^2+]

Antilog (-1.18) = (6.5 × 10-2)/[Cu^2+]

0.066= (6.5 × 10-2)/[Cu^2+]

[Cu^2+] =  (6.5 × 10-2)/0.066

[Cu^2+] =0.98 M

Learn more about Nernst equation: https://brainly.com/question/22724431

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