Question

A concentration cell is set up using the following reaction: Sn4+ (aq) + 2e – Sn2+ (aq) ° = 0.150 V The standard line notation, including initial concentrations, for this cell is shown below: (Please note that concentration of ions shown in the anode compartment are in the same order as in the cathode compartment.) Pt (s)[Sn4+] (0.645 M), [Sn2+ ](1.39 M) || (Sn4+] (1.37 M), [Sn2+] (0.611 M) |Pt (s) The initial potential of this concentration cell, at 298 K, is: & (V) = Submit Answer Tries 0/2

Answer 1

Sn2* (0.611M) >Sn" (0.645 M) + 2e Sn Reduction half reaction (cathode): Sn* (1.37 M) 2e 4+ Sn2 (1.39M) Oxidation half reaction (anode): (0.645 M) Sn2 (0.611) Sn" (1.37 M) +Sn2 (1.39M Cell reacti on: Since both electrodes of same materi al, this cell is a concentration cell and the standard cell potenti al Ewil be zero No of moles of electrons transferredin the cell, 2 = Cell reacti on: Sn"(1.37 M) +Sn2 (1.39M) According to Nernst equation Sn2 (0.611M) > Sn* (0.645 M) 0.0591V -log Q cel cel 2 (0.645) (0.611) 0.0591V =0.00 V log (1.37)(1.39) 2 0.0591V = 0.00 V -log (0.207) 2 0.0591V -(-0.684) 2 = 0.00 V = 0.00 V - (-0.0202 V) = +0.0202 V