Question

Can anyone help me check mu answer and show the solution? Thanks!

The two half cells that make up a battery are shown below with their reduction potentials: Ce4+ (aq) + e Ce 3+ (aq); E= 1.61 V S2082-(aq) + 2e – 25042-(aq); E= 2.01 V What is the equation for the galvanic cell made with these two half equations? Oa. 2Ce3+ (aq) + S2082 (aq) - 2Ce4+(aq) + 25042-(aq) Ob. 2Ce4+ (aq) + S2082 (aq) 2Ce3+ (aq) + 25042-(aq) Oc. 2Ce3+ (aq) + 25042-(aq) - 2Ce4+ (aq) + S2082-(aq) Od. None of the other options Oe. 2Ce4+ (aq) + 25042"(aq) – 2Ce3+ (aq) + S2082"(aq)

Answer 1

Your chosen answer is correct and explanation to this is explained below :
* Note it Galvanic cell is a type of Electrochemical cell where chemical energy gets converted into electrical energy. It involves two compartments consisting of half cells where oxidation reaction occurs in one compartment ie at anode while reduction occurs at cathode. For a spontaneous overall cell reaction E cell must be positive & for this we have to take that species at cathode whose retuction potential is is higher. So, As, S2 82- conversion into soy2- E is we take this at cathode ire reduction takes place, while will be taken at anode. Hence, At anode - (oxidation) (Release of electrons takes ] X2 place) it caq.) Ce et cagi tle At cathode - (Reduction) live gaining of electrons takes place) 52 og caq.) + zê - 25042- E is higher SO cest caq.) * Co 34 * 2- 2 Ce 3+ (aq.) + Sz g? caq.) - 2004 t caq.) + 2504 cag.) with, E cell = E cathode - E anode 2.01 v 1.618 = +0.40V so wrorect option is, a) 2Ce 3+ caq.) + S2 8² cag.) 2004 t caq.) + 2004 caq.)