The half reaction that occurs at the Au electrode is 1.64
<u>Explanation:</u>
Half cell reaction at 'Au' electrode
We have the equation,
Au + (aq) + e- ---> Au(s)
Given the concenration of AuNO3=0.150 M
[Au +] =0.150 m
From the equation,
Au + (aq) + e- ---> Au(s)
Standard electric potential= Eo= 1.69 volt
Solving the problem using the Nerst equation
E cell= E0 cell - 2.303 RT/ nF log Qc
Where,
T = 298 K
n= no of electron lost or gained
F= faraday's constant= 965000/mole
R= universal constant= 8.314 J/ K/ Mole
Substitue the values we get
E cell = 1.69 volt - 0.05 g/n log (1/0.150M)
E cell = 1.69 volt - 0.05 g/1 0.824
E cell= 1.64
The half reaction that occurs at the Au electrode is 1.64
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