Given the following half cells: Fe2+/Fe0 and Pd/Pd2+, determine; a) what is the anode and what is the cathode, b) what is Eo, c)what is Go and k? Using the following concentrations, determine E value for your cell. Fe2+ [.25], Pd2+ [.05]. How long would you have to run this battery at 3.2 amps to lower the E by .05 volts?
Given the following half cells: Fe2+/Fe0 and Pd/Pd2+, determine; a) what is the anode and what is...
Given the half-cells: Ni/Ni+2 and Fe/Fe+2 Chose the anode and cathode. Calculate E° and ΔG° for this reaction. Calculate what the new cell potential (E) would be if the concentrations were [Ni+2] = 0.40 M and [Fe+2] = 0.30 M. Calculate how long you would have to run the battery at 6 amps to make the E drop by 0.03 volts, and calculate how many grams of anode you lose by that time. Using the equations: ΔG° = -nFE° ln...
Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...
Consider an electrochemical cell constructed from the following half cells, linked by a KCl salt bridge. a Fe electrode in 1.0 M FeCl2 solution a Sn electrode in 1.0 M Sn(NO3)2 solution Fe2+ + 2e- → Fe Eo = -0.44 V Sn2+ + 2e- → Sn Eo = -0.14 V When the cell is running spontaneously, which choice includes only true statements and no false ones? Select one: a. The iron electrode gains mass and the iron electrode is the...
question 16.46 UU 15 - UU UL ut tu uve- 16.46 The half-cells Fe2+ (aq)|Fe(s) and O2(g)|H2O (in acid solution) are linked to create a voltaic cell. (a) Write equations for the oxidation and reduction half- reactions and for the overall cell reaction. (b) Which half-reaction occurs in the anode compart- ment and which occurs in the cathode compartment? (c) Complete the following sentences: Electrons in the external circuit flow from the — electrode to the electrode. Negative ions move...
Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...
79. Two half-cells, PtlFe3+(aq, 0.50 M), Fe2 +(aq, 1.0 x 10-5 M) and Hg (aq, 0.020 M)IHg, are constructed and then linked together to form a voltaic cell. Which electrode is the anode? What will be the potential of the voltaic cell at 298 K:
5. Consider the following set of half reactions Ag (aq)Ag (s) Cu2(a)2 e- Cu (s) Ai (aq)3eAl (s) Eo0.7996 V E0.342 V Eo-1.6632 V i. Which pair of half reactions would make the battery with the largest E cell? i. Identify the cathode and the anode for this battery, and calculate the value of E cell. i. Write the overall redox reaction that would take place in the battery above.
Consider the following electrochemical cell (battery): Mg(s) │ Mg2+(aq) ║ Cl- (aq) │ Cl2 (g) │ Pt(s) all gases 1 atm, all solutions 1.0 M a. Write the respective reduction half-reactions occurring on each side of the salt bridg, and from some reference, get the half-cell potential (in volts) for each. b. Determine what reaction occurs overall, and calculate the Eo cell for this electrochemical cell. c. Make a drawing of this cell, and label the ANODE and the CATHODE....
Given the following two half reactions with their standard reduction potentials: Rh3+(aq) + 3e− → Rh(s) E° = 0.76 V Au+(aq) + e− → Au(s) E° = 1.69 V What is the cell potential (E, in Volts) for a cell at 298.15 K whose anode consists of [Rh3+] = 0.40 M and Rh(s), and whose cathode consists of [Au+] = 0.10 M and Au(s)?
A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and (ii) Sn2+(0.236M) and Sn4+(0.145M). MnO4-+ 8H++ 5e-↔Mn2++ 4H2O Eo= 1.507 V Sn4++ 2e-↔Sn2+Eo= 0.151 V a)Calculate the galvanic cell voltage. b)Write the complete cell description. c)Write the cell reaction indicating clearly the direction of spontaneous reaction. d)Name the electrode (anode or cathode), give the polarity (positive or negative) of the electrode, and state the reaction (oxidation or reduction) occurring in the tin half-cell. Please...