Part A Calculate the pH of the cathode compartment solution if the cell emf at 298...
1. Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.610 V when [Zn2+]= 0.20 M and PH2= 0.92 atm . Express your answer using two decimal places. pH= Switch 8.76 Voltmeter 7n anode NO, Na +H (8) Cathode compartment (standard hydrogen electrode) Anode compartment NO, Zn2+ NO NO, Zn(s) —> Zn2+ (aq)+ 2e 2H+ (aq) + 2e →H(8)
Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when [Zn2+]= 0.40 M and PH2= 0.91 atm .
Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.630 V when [Zn2+]= 0.26 M and PH2= 0.99 atm . Express your answer using two decimal places
Question 9 (1 point) In the following cell, A is a standard Zn2+1Zn electrode connected to a standard hydrogen electrode. If the voltmeter reading is -0.76 V, match each electrode with its correct name. Given: Standard reduction potential of the H+/H2 and Zn2+/Zn couples are 0.00 and -0.76 V, respectively. voltmeter saltbridg -Plis) H2 (g) Zn2+|Zn electrode 1. anode standard hydrogen electrode 2. cathode
A voltaic cell is made by placing a solid iron electrode in a compartment in which the Fe2+ concentration is 2.0 x 10 -5 M and by placing a Pt electrode in the other compartment, in which the H+ concentration is 3.4 M and PH 2 = 0.700 atm. (This is a Hydrogen Electrode) What is emf (E) for the cell at the concentrations given? Fe(s) + 2H+(aq) --> Fe2+(aq) + H2(g) Hint, you do not have cell diagram, so...
Enter electrons as e A voltaic cell is constructed in which the anode is a Ni Ni2+ half cell and the cathode is a Hg|Hg2+ half cell. The half-cell compartments are connected by a salt bridge. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) The anode reaction is: + + The cathode reaction is: + V> + The net cell reaction is:...
What is the effect on the emf of the cell shown in Figure 20.9, which has the overall reaction Zn(s)+2H+(aq)Zn(s) +2 H+(aq) −→−Zn2+(aq)+H2(g),→ Zn2+(aq) + H2(g), for each of the following changes? The pressure of the H2H2 gas is increased in the cathode half-cell. Zinc nitrate is added to the anode half-cell. Sodium hydroxide is added to the cathode half-cell, decreasing [H+].[H+]. The surface area of the anode is doubled. Can someone please explain why b is emf decreases? Why...
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);...
a.) A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode (P(H2) = 1 atm) immersed in a solution of unknown [H+]. If the cell potential is 0.204 V, what is the pH of the unknown solution at 298 K? b.) An electrochemical cell is constructed in which a Cr3+(1.00 M)|Cr(s) half-cell is connected to an H3O+(aq)|H2(1 atm) half-cell with unknown H3O+ concentration. The measured cell voltage is 0.366...
In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of a Zn/Zn2+ half-cell and an H2/H+ half-cell under the following conditions: [Zn2+ ] = 0.021 M [H+ ]= 1.3 M partial pressure of H2 = 0.32 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2+ (aq) + 2e − ⟶ Zn(s) E° = − 0.76 V 2H+ (aq) + 2e − ⟶ H2(g) E° = 0.00 V We were unable...