A half-cell containing bromine and bromide ions is connected to a half-cell containing sulfur and sulfide ions to form a voltaic cell.

Question

Determine the proper direction for each half-cell reaction to assure a spontaneous, complete cell reaction. Find the value of E^°. The two half-cell reactions and their electrode potentials are as follows:

Br₂ + 2e^- ⇔ 2Br^- E^º = +1.06v

S + 2e^- ⇔ S^-2 E^° = -0.51v

Thank you :)

Answer 1

S^2- | S || Br₂| Br^1-

E^o_{cell =} E^o_cathode - E^o_anode = 1.06v - (-0.51v) = 1.57v

As the E^o_cell is positive, the reaction is spontaneous and proceeds as follows:

S^2- + Br₂ ======> 2Br^1- + S