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4. What is the cell potential for Mn Mn2||Co2|Co, based on the standard potentials? (2 pts) Mn(s) +Co+2 (aq) → Co(s) + Mn+2(aq) 5. A Student constructed a Mg Mg2||Ag|Ag cell and measured a cell potential of 3.17 V. Calculate the cell potential of the Mg Mg2 half reaction. (2 pts)

Answer 1

- O CO = + 1:16 V 2 ē (aq) 4) Standard cell potential, Fella Kathadet Eanode - Ereduction + Eoxidation For the reaction, Mn1MD2+/1 6027100, the 2 half reactions are, Mn (3) —> Mnatcaq) + 2ē Foxiclation corean) +27 - (03 ; Edduction = – 0:28 V Net reaction : Mng + correto Minima) + Cocos Fceu = Foxidation + Eqreduction ang = +1:18 V) + (-0.28V) = +0.9V * Please note, Mnin+2ě —— Mnco, es given is reduction (219) and gº for this reaction is - 1.18 V. If the eceaction is reversed, it becomes oxidation & therefove olgn of Eº also changes. MNCs) → Mn2t+2ě ; 5° = +1.18V Ics Scanned with CamScanner

5) Mg/ Mg2+ || Ag+ lag Feriell = 3-17 V 2 half meections are, Mg — Mg24+ 2ē ; Eoxication = ? '2Ag+ + 2ē —> 219 ; Erecluction = +0.8V Net reaction : M909) + 2Agen) Mgratis + 2 Ag Foelle = Empeduction + Foxidation 3:17 V 3-170=0.8V + E oxidation Foxidation = 2:37 V Cell potential for Mg/ Mg2+ half reaction- Scanned with CamScanner

6) 2 Ag Cs) + 2 H*(aq) + 2 Agcaq) + H2g) 2 AJ6) 2Ag+ + 2ē ; Femidation-0-$V 2 H Taq) + 2ē → Hz gg) ; Ereduction = 0.00 .: Ecell = Freduction + Foxidation = -0.8V For a reaction to be spoznat spontaneous, AG (gibbs gelee energy) must be negative. 46 = -n Freell n=no. of electrons transferred F→ Faradays constant = 96485 /mol here, 1G = -2x 96485 x (-0.8) = 154376J 4G>0 This meaction is Non-Spontaneous Scanned with CamScanner

7) In voltaic cell, anode is considered negative & cathode is considered positive In a electrolytic cell, anode is taken to be positive & cathode as negative Scanned with CamScanner