Question

2. You measured cell potentials (voltages) for reactions of metals and metal ions. We will use the standard date and practice calculating these. Recall: E-Ecathode - Eanode Remember reduction occurs at the cathode; oxidation occurs at the anode For the anode reactions, your reactant will be on the product side of the standard equation, but don't change any sign. The equation does that for you. Cathode Reaction Agt + e → Ag Ecathode Anode Reaction Cu → Cu? + 2e Eanode Reaction Cu + Ag* Ag + Cu2 Zn + Cu2+ Al + Cu2+ Zn + A13+ Fe + Cu2 When Eº is positive, the reaction spontaneous. Circle the reactions that are spontaneous. Choose three other metals, write the half-reactions between them and decide on the order of their oxidizing strength. Reaction Cathode Reaction Ecathode Anode Reaction Eanode
please answer the full question.

Answer 1

E = Ecathoden - Eanode. first we have to know wheather the vaines are no meant for all, IM that is sooo s Ecurt cu = 0.34 v 5 Asting boor 3 Ast/Ag - o-7996. soientlan 2-0.76. etiboristeri ats A13t|A = -1.66 y at & reztlfe =-0.44 Ag teu 2t 37 Entcut 47 Al & cu²t Ental 37 6retcy t Reaction cathoderanf Ecathode Teut Ag+ Agtte tag 0-7996 Leute Ieu httze an 0.34 eukt tze all 0.34 eurthe-en 10.34 A13t+Besal 1-1.66 custthecy 10.34 Anode ran. I SAnode cuscuze 0.34 Aga Agte 0.7996 En zuze-0.76 Alast 3 -1.66 enzure 0.76 re-retre -0.44

I cell. E cathode Anode F 0.7996-0:34 20:46 0.34 -0.7996 = -0.40 6.34 to 76 = 1 L - Een positive so spontenious. non spontenious (E =-ve). spontenions. spon te nious. 10.34 +166 = 2 -1.66-6-76 .3 - 10.34 +0.44 = 0.788) ron spontenious. spontenious. Remember, whose red potential value will be greater that specites will always a oxidise other species.