The answer is +0.227 I just don't understand how to get that.
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The answer is +0.227 I just don't understand how to get that. Given the information below,...
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) |...
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.
Right side (- Keep each reaction as a reduction (as in Table), and use this for both sides: E-E- ...
right side (- Keep each reaction as a reduction (as in Table), and use this for both sides: E-E- (RT/n F) In Q Then use: Ecl E, (on the right side) - E. (on left side) 5) Consider the following electrochemical cell. Ni (s) I NİSO, (aq) (NiSO4-0.0020 M) Il (Cu2+-: 0.0030 M) | Cu (s) CuCl2 (aq) a) Write out the Nernst equation for both of the half reactions. Calculate the potential (voltage) for the system (E cell and...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...
QUESTION 4 Using the table below and the Nearnst Equation E-E° (0.0592/n)logQ Find the voltage of...
QUESTION 4 Using the table below and the Nearnst Equation E-E° (0.0592/n)logQ Find the voltage of the cell: Cu2+(aq) + Ni(s) ? Cu(s) + Ni2+(aq) at 25° C if the concentrations of the soluble species are [Cu2+1-0.050 M and [Ni]- 1.40 M Reduction Reaction Sn2++ 2e-? Sn Cu2+ + 2e- ?Cu" Reduction Potential, Ecell 0.14 Volts 0.34 Volts 0.25 Volts +0.77 Volts Fe3+ + e-? Fe2+ 0.50 V 0.54 V 0.62 V 0.66 V
A galvanic cell based on the following reactions Cu 2+ (aq) + 2e- + Cu(s) E°=0.339...
A galvanic cell based on the following reactions Cu 2+ (aq) + 2e- + Cu(s) E°=0.339 V (AD + 14H(aq) + 6e-4 2Cr 3+(aq) + 7 H2O(1) Eo=1.330 V a) Write the overall cell reaction and determine its voltage. b) If the E value of the galvanic cell is 1.2 le of the galvanic cell is 1.254 V, calculate the pH of the cell when [Cu2+]=0.00010 M, Cr-0,2-1=0.00460 M, Cr3+1=1.0x102 M. The Nerst equation is E=E - (0.0592/n)logQ
Given the following standard reduction potentials choose the cell which will work as a voltaic cell....
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Using the information in the table: Which combination of metals, if used to create an electrochemical...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
Given the following answer the questions below: a) the metal that is the cathode is ________...
Given the following answer the questions below: a) the metal that is the cathode is ________ and the metal that is the anodes _______ b) reduction takes place at the Ag or Cu electrode (circle one) c) Oxidation takes place at the Ag or Cu electrode (circle one) d) Identify the salt bridge on the cell diagram. e) Calculate Ecell: Cu2+ (aq) + 2e- --> Cu(s) E= +0.34 V Ag+(aq) + e- --> Ag(s) E= +0.80 V
please show work and how they got the answer 13. If [Cu2+] = 0.30 M for...
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,...
Calculate the cell potential for the reaction below when the concentrations of ions are: [Ag+ ]...
Calculate the cell potential for the reaction below when the concentrations of ions are: [Ag+ ] = 0.010 M and [Cu2+] = 0.750 M, at 25 °C. Cu(s) + 2Ag+ (aq) --> Cu2+(aq) + 2Ag(s) Given the standard reduction potentials: Cu2+(aq) + 2e– → Cu(s) Eϴ = 0.34 V Ag+ (aq) + e– → Ag(s) Eϴ = 0.80 V (A) 0.35 V (B) 0.44 V (C) 0.46 V (D) 0.48 V (E) 0.57 V
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.
right side (- Keep each reaction as a reduction (as in Table), and use this for both sides: E-E- (RT/n F) In Q Then use: Ecl E, (on the right side) - E. (on left side) 5) Consider the following electrochemical cell. Ni (s) I NİSO, (aq) (NiSO4-0.0020 M) Il (Cu2+-: 0.0030 M) | Cu (s) CuCl2 (aq) a) Write out the Nernst equation for both of the half reactions. Calculate the potential (voltage) for the system (E cell and...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...
QUESTION 4 Using the table below and the Nearnst Equation E-E° (0.0592/n)logQ Find the voltage of the cell: Cu2+(aq) + Ni(s) ? Cu(s) + Ni2+(aq) at 25° C if the concentrations of the soluble species are [Cu2+1-0.050 M and [Ni]- 1.40 M Reduction Reaction Sn2++ 2e-? Sn Cu2+ + 2e- ?Cu" Reduction Potential, Ecell 0.14 Volts 0.34 Volts 0.25 Volts +0.77 Volts Fe3+ + e-? Fe2+ 0.50 V 0.54 V 0.62 V 0.66 V
A galvanic cell based on the following reactions Cu 2+ (aq) + 2e- + Cu(s) E°=0.339 V (AD + 14H(aq) + 6e-4 2Cr 3+(aq) + 7 H2O(1) Eo=1.330 V a) Write the overall cell reaction and determine its voltage. b) If the E value of the galvanic cell is 1.2 le of the galvanic cell is 1.254 V, calculate the pH of the cell when [Cu2+]=0.00010 M, Cr-0,2-1=0.00460 M, Cr3+1=1.0x102 M. The Nerst equation is E=E - (0.0592/n)logQ
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
Using the information in the table:
Which combination of metals, if used to create an
electrochemical cell, would produce the largest voltage?
Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...
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,...
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Amplitude=3;
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fftSignal= fft(signal);
fftSignal=f ftshift (fftSignal);
f=fs/2*linspace(-1,1,fs);
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xlabel('Frequency(Hz)’)
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