The of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵
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
E° = -0.403 V
At the cathode, reduction occurs
E° = 0.34 V
Overall reaction:
We know
= 0.34 -(-0.403) = 0.743 V
Also,
ΔG° = -nF
Where, n = no of electrons gained or lost
F = Faraday constant
= 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 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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