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please show work and how they got the answer 13. If [Cu2+] = 0.30 M for...
please show work and how they got this answer 12. Consider the cell reaction Cu(s) + 2Agt → Cu2+ + 2Ag(s). If [Ag+] = 1.0M and Ecell = 0.62 V, then what is the concentration of Cu2+7 Eºcell = +0.46 (0.80 -0.34), Q = x/1.02, x = [Cu2+] = 3.9x10-6
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)
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.
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) +...
Conceptual: Consider a Sn(s)|Sn2+(aq) || Cu2+(aq)|Cu(s) cell. If the Sn2+ concentration is increased, what will happen to the measured Ecell value? • Calculation, full-reaction Nernst equation: Use the full Nernst equation to calculate Ecell for the conditions described… • Easier: Ni | Ni2+(0.300 M) || Cu2+(0.002 M) | Cu • Harder: Al | Al3+(0.002 M) || Cu2+(4.00 M) | Cu • Calculation, half-reaction Nernst equation: Use the Nernst equation to calculate E at pH 3.00 and [Cl- ] = 0.0035...
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
The % difference = Standard E° (using table) = (show calculation) Ee=Fe+2/ Fe= -0.44V Eo=Cu2+/ Cu= +0.34 V Eo cell=Eocathode-Eganode = Ee Cu+2/cu- Eo Fe2+/ Fe = +0.34+ 0.44 V = +0.78 V Note that the list of reduction reactions includes the reduction of ironfil) to elemental iron, and also the reduction of iron(III) to iron(II). Is your data more consistent with the formation of Fe? or Felt? Explain your answer.
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.
I need help with questione 1-12 and discussion question 1 and 2. The previous pictures help determine the chart. Please Show Work thank you so much An oxidation half-reaction is characterized by electrons appearing on the product side. The oxidation of aluminum for instance would be represented thusly: Al(s) → Al3+ + 3e- (1) An reduction half-reaction is characterized by electrons appearing on the reactant side. The reduction of ferrous iron for instance would be represented thusly: Fe2+ + 2e...
7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s) Fe2+(aq) (1.1 M) || Cu2+ (aq) (0.50 M) Cu(s) Ecell = Eºcell - 0.0592/n logQ