Question

A voltaic cell has one half-cell with a Cu bar in a 1.00 M Cu²⁺ salt, and the other half-cell with a Cd bar in the same volume of a 1.00 M Cd²⁺ salt.(a) Find E°cell ΔG°, and K.

Answer 1

The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

What is a voltaic cell?

A voltaic cell often called a galvanic cell, is an electrochemical device that produces electricity through spontaneous redox processes.

It is divided into two distinct half-cells. A half-cell is made up of an electrode (a metal strip, M) dissolved in a solution containing Mn⁺ ions. M can be any metal.

A wire from one electrode to the other connects the two half-cells. Additionally, a salt bridge links the two half-cells.

To solve the question, we need to write the equations of two half-cells

At the anode, oxidation occurs

$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$

E° = -0.403 V

At the cathode, reduction occurs

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

E° = 0.34 V

Overall reaction:

$Cd(s) + Cu^{2+}(aq) \rightarrow Cd^{2+}(aq) + Cu(s)$

We know

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

= 0.34 -(-0.403) = 0.743 V

Also,

ΔG° = -nF$E^o_{cell}$

Where, n = no of electrons gained or lost

F = Faraday constant

$E^o_{cell}$ = standard potential

ΔG° = -2×96485×0.743 = -143376.71 J = -143.377 kJ

Also,

ΔG° = -RTlnK

-143.377 = -8.314 × 10-3 × 298 × lnK

lnK = 57.87

K = 1.35 × 10²⁵

Hence, The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

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