3. For the reaction below, what is the value of E when [Cu2+]-0.04 M and [Fe2+]=0.4...
1- Consider the following redox reaction: Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s) EoCell = 0.78 V If [Cu2+] = 0.3 M, what [Fe2+] is needed so that Ecell = 0.76 V 2- Calculate the Eocell for the following redox reaction: Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s)
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at the cathode and what is the Eceu? Fe2+(aq) + 2e → Fe(s) E°=-0.44 V Cu²+(aq) + 2e → Cu(s) E°= 0.34 V a) b) c) d) Cu2+(aq) + 2e → Cu(s), Ecell = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecel = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecell =-0.10 V Cu²+(aq) + 2e → Cu(s), Ecel = 0.10 V Cu²+ (aq) +...
please show work and how they got the answer 13. If [Cu2+] = 0.30 M for the same cell described below, then what concentration of Fe2+ is needed to result in Ecell = 0.76 V? You will need to use the standard reduction potential table. Cu2+/ Cu → E = +0.34 V Fe2+/ Fe → E = -0.45 Cu?* (aq) + Fe(s) - Fe* (aq) + Cu(s) Eºcell = +0.78 V, Q = x/ 0.30, X = [Fe2+], Using Nernst,...
Consider the following reaction at 25 °C; Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+ (aq); E0cell = 0.78 V (25 °C) What would be the value of Ecell at 25 °C, if [Fe2+] = 0.40 M and [Cu2+] = 0.040 M.
What is the OH ion concentration in a 4.8 x 10-2 M KOH solution? 2.1 x 10-13 M 4.8 x 10-2 M 1.0 x 10-7 M 4.8 x 10-12 M What is E' for the following balanced reaction? Fe(s) + Cu2+(aq) Fe2+ (aq) + Cu(s) Standard Reduction Potential Half-reaction Fe2+ (aq) + 2e Fe(s) Cu2+ (aq) + 2e Cu(s) -0.44 40.34 O +.010 +0.78 0 -0.78 -0.1
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
9) A sol'n in an electrolytic cell contains the ions Cu2+, Fe2+ & Zn2+. If the voltage (initially very low) is turned up, in which order will the metals be plated out onto the cathode? Cu2+(aq)2e-Cu(s); E0.34V Fe"(aq) + 2e-→ Fe(s);E"--0.44V Zn2+(aq) + 2e-→ Zn(s): E。--0.76V 10) How long will it take to produce a sol'n of pH 2.00 by electrolysis of 500 mL of 0.100 M AgN03(aq) at 0.240 Amperes? Anode: 2H2O(l) → 4H-(aq) + O2(g) + 4e-, E°...
26. Determine the redox reaction represented by the following cell notation. Fe(s) I Fe2+(aq) I| Cu2+(aq) I Cu(s) A) Cu(s) + Fe2+(aq) B) Fe(s)+Cu2+(aq) Cu(s)+ Fe2+(aq) C) 2 Fe(s)+ Cu2+(aq)Cu(s) + 2 Fe2+(aq) D) 2 Cu(s) + Fe2+(aq)Fe(s)+ 2 Cu2+(aq) E) 3 Fe(s) + 2 Cu2+(aq) 2 Cu(s)+3 Fe2+(aq) Fe(s) + Cu2+(aq)
I have no clue on how to do this, please help. Consider the following reaction: Cu2+ (aq) + Fe(s) → Cu(s) + Fe2+ (aq) AE° = 0.78 Volts Calculate the cell potential (AE) when the concentration of Cu2+ is 0.040 M and Fe2+ is 0.40 M.
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С