Question

A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.170 M AuNO, 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. half-reaction: The standard reduction potential for the half reaction that occurs at the gold clectrode is E 1.69 V. What is the potential cf the half-reaction that occurs at the gold electrode under the given conditions? Ignore activities. E= The reduction potential for t the saturated calomel electrode is E-0.241 V. What is the cell potential?

Answer 1

In potentiometry , reference electrode is connected to negative terminal and indicator electrode is connected to positive terminal of battery.

Reference electrode acts as a anode and indicator electrode acts as a cathode.

Here,Reference electrode is SCE and indicator electrode is Au electrode.

Reduction reaction will occur at Au electrode.

Au ³⁺_(aq) + 3 e^- $\rightarrow$ Au _(s)  

Here ,n = 3 = no of electrons

Potential of reaction taking place at Au electrode is

E = E ⁰ red - (0.0591 /n ) log 1/ [Au ³⁺ ]

= 1.69 - (0.0591/3) log 1/ 0.170

= 1.69 -( 0.0591/3 ) 0.7696

= 1.69 - 0.01516

E = 1.675 V

Cell potential is calculated as E cell = E cathode - E anode

= E _{Au3+ IAu} - E _SCE

= 1.675 V - 0.241 V

   E cell = 1.434 V