the potential between Cu and the unknown is .98 (some of the pictures may be unneccesary, but wanted to provide all details)

Part C: Determine the E for a Voltaic Cell Using Cu and an Unknown Metal C1. Use equation (5) from the Lab 10 - Background to solve for E anode of the unknown metal, keeping in mind that the Cu is the cathode for the cell when the Ecell is positive. Show all work. C2. Compare the E anode of the unknown to the values listed in Table 10-1 (shown above) and state: "The unknown metal is Cu NO 31 Fe(NO3)3 Fe O Cu (Nels) 97 Zn (NO3) O Zn(NO3) 2NDS ( Zn Zn M (NO3) PART B POST LAB B2) Cu (s)-Zn(s) - Half-cell-Rxn (cathode> Half cell Rxn 2n(s) Cu2+ 2e Zn+2e QUESTION BA half Anode reaction poternal Reduction QUESTION A0, A1, A2 Cu (s) fe (s) Zn (s) A3.) Cu (NO)2 A1 A2 NO Fe (NO) A1 A2 NR No NO NR 2+ Fe(s) +lu (aq) Cu (s) + NR .81 Zn(s) + Cu2+ +Z 1. 10V Fe+ (aq) +2 + Cu + Zn+ 2+ Fe(s) + znzt (agl . a Zn (NO3)2 A1 NR NR NR Reduction METAL / 2+ 24 Zn/z2 A2 No NO NO POTENTIAL EV -34 -.76 -.47 Electrode E Ag++eAg +0.80 V Cu2+ + 2e +0.34 V Cu Pb2+ + 2e -0.13 V Pb Fe2+ + 2e -0.44 V Fe Zn2+ + 2e -0.76 V Zn Al3+ + 3e -1.66 V Al Mg2+ + 2e -2.37 V Mg