Question

7. Electrochemistry and the Nernst Equation Keall Leaure thbala Name Section PreLab Questions Calculate the value of E°cell for the following reaction: Au(s) +3Ca2+(aq) 2AU3+(aq) + 3Ca(s) 1. 2. Calculate E°cell for a silver-aluminum cell in which the cell reaction is Al(s)+3Ag+(aq) A13+(aq)+3Ag(s)

Answer 1

m Reduction potential value! Au² (aq) + 3e - Auc) car caq.) + 2e - Cacs) Apaq) + 38 = Alis) Agt caq.) + e ² Ag (8) = +1.50 v & -2.87 v = -4.66 v E = +0.80 v 2 ALL (5) T 9la caq.) 37 2 Acu caalt 3 Cals) Break up the overall meaction into halk reactions oxidation : 2 Au(s) → 2 Alestaan + be Reduction: 3 Cottag.) + te Cars) Oxidation potential of Au/A43t is same as the reduction potential of Außt / Ace, but with opposite sign. we can find Ecell as follows. Feelis Eoxidation + Ereduction Bien : -(1.50) v --- 2.87 V = -4.37 V At we know that in which oxidation Becurs that electrode is anode and in which reduction occurs that electrode in cathode, So, we can write the above equation as E = ( E8x d' Anade & (& Rear cathode

Ecen= (-Exeir Anagle+ (East Cashode Ecen - (Ered" enthode (Exedro) Anock e Cen) = -2,871 – 1.50V. 1.SOV 3 -4.39 V

Another thing to remember is that higher reduction potential material is actually oxidize a to reduce the lower reduction potential material a more positive the reduction potential, the greater the species' affinity for electrons and tendency to be reduced. Here. Exidatien = oxidation potential t Reduction & Reduction potential However , E means to reduction for simplicity in electrochemistry, so instead of writing erred we can corite E.

( 2 A l(s) + 3 Agtca - A Cag) + 3 Ages Half reaetions oxidation: Ales) — Altant 300 Reduction : 3 Agt eag.) → 3 Ag (y) .: + 360 Ecell - É cafrode - {Anode = +0,80 V - (-1. 66 V) € 2.46 V Note: Eren co, then the reaction is not spontaneous I and it will not occur as wroetten under Standard condition, it will , however, proceed spontaneously in the opposite direction. Ecell > Ö means that the reaction will occur spontaneously as written...