Determine the cell notation for the redox reaction given below. Sn(s) + 2H+(aq) ⟶ Sn2+(aq) + H2(g)

a. H+(aq) | H2(g) | Pt ∥ Sn(s) | Sn2+(aq)
b. H2(g) | H+(aq) | Pt ∥ Sn2+(aq) | Sn(s)
c. Sn2+(aq) | Sn(s) ∥ H2(g) | H+(aq) | Pt
d. Sn(s) | Sn2+(aq) ∥ H+(aq) | H2(g) | Pt
e. Sn(s) | H2(g) ∥ Sn2+(aq) | H+(aq) | Pt

The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Explanation:

The half reactions are:

2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)

Sn(s) ⟶ Sn²⁺(aq) + 2 e-  (oxidation)

In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:

anode reaction∥ cathode reaction

where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.  

In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).

The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:

Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

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