A cell consists of a gold wire and a saturated calomel electrode (S. C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire is connected to the positive end of a potentiometer, and the S. C.E. is connected to the negative end of the potentiometer. What is the half‑reaction that occurs at the Au electrode? Include physical states.

The half reaction that occurs at the Au electrode is  1.64

Explanation:

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

the answer is definitely c.