Question

Consider a circuit in which the left half-cell was prepared by dipping a Pt wire in a beaker containing an equimolar mixture of Cr2+ and Cr3+. The right half-cell contained a Tl rod immersed in1.00 MTlClO4.

(a) Usetheline notation to describe this cell.
(b) Calculate the cell voltage.
(c) Write the spontaneous net cell reaction.
(d) When the two electrodes are connected by a salt bridge and a wire, which terminal (Pt or Tl) will be the anode?

Answer 1

a) Ti⁻²|Ti || Pt|Pt⁻²

b) +2.83 V

c) Pt + Ti⁻² → Pt⁻² + Ti

d) Pt

Step-by-step explanation:

a) In the cell circuit, a redox reaction will happen, and one of the electrodes will oxide (it will lose electrons) and the other will reduce (it will gain electrons). The potential to reduce is measured by the standard potential reduction (E°). As higher is it, as easy is to the reduction reaction happens. The values for Pt and Ti are:

Pt + 2e⁻ → Pt⁻² E° = +1.20 V

Ti + 2e⁻ → Ti⁻² E° = -1.63 V

So, Pt will reduce and Ti will oxide. The oxidation reaction is the opposite of the reduction, and at the line notation, first is placed the oxidation reaction, thus:

Ti⁻²|Ti || Pt|Pt⁻²

b) The cell voltage (ΔE°) is the E° of the reduction reaction less the E° of the oxidation reaction:

ΔE° = 1.20 - (-1.63)

ΔE° = +2.83 V

c) The spontaneous net cell reaction will be the sum of the reduction reaction and the oxidation:

Pt + 2e⁻ → Pt⁻² (reduction)

Ti⁻² → Ti + 2e⁻ (oxidation)

----------------------------

Pt + Ti⁻² → Pt⁻² + Ti

d) In the cell, the electrodes are called cathode and anode. The electrode where the oxidation occurs is the anode, and the other, where the reduction occurs, is the cathode. Thus, in this case, the Pt electrode is the anode.