Question

need help with the rest of the table

EXPERIMENT 10 DETERMINATION OF THE ELECTROCHEMICAL SERIES PURPOSE To determine the standard cell potential values of several electrochemical coll INTRODUCTION The basis for an electrochemical cell is an oxidation reduction Corredor be divided into two half reactions reaction. This reaction can Oxidation half reaction Gloss of electrons) takes place at the anode, which is the positive electrode that the anions migrate to Chence the name anode) Reduction half reaction (gain of electrons) takes place at the cathode, which is the negative electrode that the cations migrate to thence the name cathodel Due to the difference in the electric potential between the two electrodes, an electrical current can be generated. This current can be used for many purposes and is the basis of all battery technologies The difference in potential is a result of the differences between the individual potentials of the metal electrodes with respect to the electrolyte, which is an electrically conductive jonie solution. The electric potential also varies with temperature, concentration of electrolyte, and pressure. The standard electrode potential, E, is the measure of potential of any electrode at standard conditions temperature at 299 K, concentrations at 1 M and partial pressures of gases at 1 atm). Since the oxidation potential of a half reaction is the negative of the reduction potential in a redox reaction, it is adequate to calculate either one of the potentials. The standard electrode potentialis commonly written as the standard reduction potential In this experiment, the potentials of several different metals will be compared by measuring combinations of their half reactions. Standard potential is defined as the potential of an electrode that consists of a metal and the 1M solution of the salt of that metal. The representation of an electrode is always "oxidation/reduction." For example, Zn/Zn represents a zinc electrode. If the solution portion of two electrodes is connected electrically, we have a galvanic cell. The potential measured between the two metals is the electromotive force. In one electrode, spontaneous reduction occurs in the other, spontaneous oxidation occurs. For example, if a 2nZn electrode and a Cut/Cu electrode are connected, copper will oxidize zinc since the zinc is located above the copper (that is, zinc has lower reduction potential than copper, according to the table below). In the table below, metals are arranged in increasing reduction (or decreasing oxidation) potential from top to bottom. That is, for a metal further down in the table, the greater its reduction potential is since it is easier to reduce it. The opposite is true as well: the higher up a metal is in the table, the greater its "oxidation potential," since it is easier to oxidize it. For instance metals located above the copper are expected to reduce copper (since for those, copper has the highest reduction potential). Metals located below the copper have higher reduction potential Therefore, those will be reduced and the copper will be oxidized. EXPERIMENT 10 11

Answer 1

I am trying to solve this using values given here.

Solution + 2+ At cathode - cu2tte Cuis) E = 0.34 (Reduction portent _ AGZ - NEED 2-2 F10:39) At anode - Fe Fe + 2e E = 0.99 Coxidation) boledat AGů za UFE We can not direct substrate anode scaduction potential from cathode it is sometime give wrong value (but how we com do) electrode potential is not additive - AG Halal = AG Abiz AGitolal z - 2F(0-34) = 2XFX(0.44) 1 -NEAE 2 -2F1031) – 2 FX(0.99) SFAÇ° -(0434 +0.99) AE = .789 Hence of erros a Actual - Calcolated 1 Xtoo on Actual of eXM091 z 0-78 – 0.65 x 1m 0.78 - 0.130 X{co 0.78 01667 X 100 z 16.6744 Hena End (Cathode) - 0.34 v Erud (Anode) z - 0.999 calculated reelt z_ 0.78 V do Error = 16 671 POLO

7 DATE: 1 PAGE NO.: e at cathode - Agt te" + Agos) Escud z o.fov at anode - Fecs) Feet +2e Eoxid a 0.94 v 2Agt + Fecs) eggs) + Feet DEʻz 0.80 +0.44 z 1-24 volt Here we can add eredueffon potential face simplicity because it makes cell. Otherwise reduction potential is not additive Henel 7. Eriral z 1.24 - 103 . -X Low z 2. 1oxto you 124 7.24 0.2 0.210 x100 3 1-29 2 16.93% Неии NP Ered (cathode) zo.80V Feed (Anode) = -0.490 calculated Erell z 1.24 V do Tehron z t6.934.

0 PAGE NO.: 2% at cathode - Felt +2€ → Fels)_Erud 2 -0.94 V at anode - Mgces + Mght fze" Eomad z_+237v Fe$t +mges) + Fecs) + Mglet Bell z_ -0.897_+2+371 z 1:93 v | CAMOM z 1-93-7:44 x 16o z 40.936/- 1-93 Henel E sead (cathode) z -0-441 Eried Cauode) z -2.37v Dorell z 4-932 tevrou z 40934. at cathode Sh*+2e + Sn(s) Esed 3 -0.13 V _ at anode Flis Feet the Eoxd z 0.49V snt + Fecs) - Sncs) + Fet Erell Z 0 13V+0.44Vz0:3100v of Cremony z 0.31.00 -0.05 x 100 = 83.877. 6.3100 Exed Ccathode) 2-0:13v Egund Canode) = -0.99 V Earl 7 0 3100 V /. peror z 83.87% POLO

1 DATE: PAGE NO.: 15 at cathode - __$$$*+_3e" + Hlcs Escad 2 -166V at anode - Fe + FePt+2é" ent Exz0.980 © AGZ-FE Multiplying with 2 and 2 with 3 then add 2483+ + 3 Fecs) Y ZAICS) +3722+ - 6 PAESeth e 21-3EXE=1:66)) +3(-2F (0.49)) Arcell z - 166 +0.44 = -1.220 7 euros z 122-007 nie I X 100 z 99264 T22 Hence- Exed (cathode) z -1.667 Esed Cauode) = -0.94V Ecell-=_-1-22V o vorost z 94-26%. at cathode - Fe 2+ tzÈ Fe (s) Erud zoo.gov at anode - - Zn → Zn²+ + 2€ Ecxd z 0.76V Fe* + + Fecs) + Zn2+ Enllz -0.44 +0:76 z 0.3200 V • Dvou z -0.32000-29 — 0.3200 z 9.37% Exed (cathode) z -0.44V Esed Canode) z 0.76V Ecell a 0.32001 , terrou z 9.37% POLO

1 DATE: PAGE NO.: 0 - at cathode - Agt te Ages) Euda 0.80V at anode - - Cu Cu²+ + z ł Fox = -0.34 v multiply 1 with 2 and adding we get 24gt +Coce) +Agcs) + cute »--2F Ecell 29f-1 F (0.80)) + (-2F(-0.34)) Z 0.80 -0.39 0.46 V eta errey z 0.46 – 0-17 x 100 046 _263.09972 Herce Ered (cathode) z 0.800 Fred Canode) z o.34V Ecell 2 0.469 do exos z 63.0940