Question

Chem 1212 Lab Report on electrochemistry

Electrochemistry When electrons transfer between reaction components in a redox reaction, we can harness the motion of the electrons to create a potential. Electrochemistry revolves around the separation of the two half-reactions in a redox reaction and establishing two different electrodes. This might involve physically separating the half-reactions or including a separator, such as a semi-permeable membrane or plastic dividers. With the reactions separated, the electrons will need to flow through the wire connecting the electrodes, creating a current. The spontaneous flow of electrons in this set-up is known as a voltaic or galvanic cell. This is the basic idea behind a battery. The cell generates electrical energy as the electrons spontaneously move from the negative electrode to the positive electrode. Voltaic Cells A cell with a spontaneous flow of electrons is known as a voltaic or galvanic cell, with the set-up as shown in Figure 1. The negative electrode is where the oxidation occurs and is known as the anode. Because oxidation is the loss of electrons, the anode is where electrons flow from. In the voltaic cell shown, the beaker with the zinc metal and zinc solution is the anode. Electrons flow toward the cathode where reduction occurs. In this cell, the beaker with the copper ions and copper metal is the cathode. SHEIN 2020 Direction of electron flow NO, K Zn → Zn2+ + 2e Cut + 2e Cu

Answer 1

Zo?? + 2e → Zn(s): -0.763 X Voltaic cells A Fera Frathad - Førode. RP faello Feel ang, fem - folk-0.0594 3. ng g; oostinga 2n + Curet 2n?++ curs) [221] [cui (2011) [cult] lomas [{n?! Jull [C121] Pb +2 mg * —- PB?**?ng Half Reaction 8 (ul): 2 Ag+ + e - Agli) 70.799 Ca2+ + 20 Cels) Pb2+ + 2e + Pb(s),-0.126 +0.337 a = pood [Ag+]2 Cell Fall lu) Reaction Quotient. 727+(1.0M) g- Fall (W) = 0.337-60.16) - E- 0.059 Lang g 31|2^2*(2.0%) 11 (174/14 4 Lust (2.0M) = 1.4 V 2 ? 1.1 -0059 (2 f 1,1 v (n=2 -0.0594 Zn|7=460) || (v (0-com) fue 85 2402) 1.1. V 1.1 -0.0 Cep 24 (0.10 M) = 1.07 v. 20123201))) ca 2cm) (1/3) Zo?*(0.2M) =2.1 V. 0= (42+ (1.0M) 1.2-0.059100(0-2) :1-0,9590g = 1.12 V PbID] pb2*(pen)|| Ag(09)pg 63- Pb2+(1.01) = 0·799-C126) [ng] C2.0) -0.925 V. -E-0.059 1.0 g 01.02 = 0.925- 0.925 V. Pb/Pb/803/199*! ng = 0.925 6 = Pb²(1.01) [Ag +J" (1) 0-01 = 0.866 v, 16 Pb|p6*600)|| Ag*o9] mg 10 - Pb2+ (0-10) Ag10) 0:925-09.win oosing - 0.925 = 0.925-0.0596 0.1 -0.954 v. Full remain same CSScanned with CamScanner