Question

A voltaic cell utilizes the following reaction and operates at 298 K.

3 Ce4+(aq) + Cr(s) 3 Ce3+(aq) + Cr3+(aq)

(a) What is the emf of this cell under standard conditions?_____ V

(b) What is the emf of this cell when [Ce4+] = 1.3 M, [Ce3+] = 0.013 M, and [Cr3+] = 0.014 M?_____ V

(c) What is the emf of the cell when [Ce4+] = 0.51 M, [Ce3+] = 0.88 M, and [Cr3+] = 1.2 M? _____V

Answer 1

Given. Reaction Reduction - af 3ce4 + (s) -3 [etmeyt G $T(94) L oxidation Half cell Reaction oxidation cr(s) – 36 —Crot (aq) • fo Reduction : Centralt de - Ce 3+ (aq) - Mentiply (3) to ean (2) to make electron similary we get € Ecratfor other (③ ECA/C = 14 4 + ces can Ecen a EceAlcast - Eestler Add equo & equ 3 to get net reaction, er(s) +3ce4+ (+(an) 12€ 3(aq) cell :- Pt .|cr|cv3+|| ce3t|cettfet Eceu a Esight - Eleft Ecem = {144 - (0.74) Žv Freie = (*: 44 +0%) v = 12:18 v/ y Nexst Eqn ał 298* Econa Econ - 0.059 log [ as erst Fins a 2. 18V- og log Conificoruz cs Ecelea 2:49 V -0.059 log (1. 4X1078) = 2.18v – 0959 * (* na 3 ho ole transferred giren [оезv): о-озм (ex3) = 0.0149 [c247] = 1:3M

2:18V +01547 = 12.334 v1 Giver leat - 2M Eceu E en . 0.054 16 osim es Ecell EU 0.959 log [ceny jes") 52:18 - onedlig 100* (4.2) 5 2.18 -0.057 by cut Tentjs 2.18 -0.059 l 2.18 -0.05 3 Ecen = 2.18 – 0,059 (0-10) Ecell =(2.18 - 0.0155 v 3 cs Scanned with = 2º164577 CamScanner is Str

a) the emf of this cell under standard conditions : 2.18V

(b) the emf of this cell when [Ce⁴⁺] = 1.3 M, [Ce³⁺] = 0.013 M, and [Cr³⁺] = 0.014 M =2.334 V

(c) the emf of the cell when [Ce⁴⁺] = 0.51 M, [Ce³⁺] = 0.88 M, and [Cr³⁺] = 1.2 M =2.1645 V