
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
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 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...
A. Calculate the cell potential that would result in the following electrochemical cell at 25oC: Pt | UO22+(0.10M),U4+(0.22M),H+(0.60M) || Ag+(0.034M) | Ag Line notation is anode || cathode Write Nernst equations for the anode and cathode separately. Then Ecell = Ecathode - E anode B. If the electrochemical cell Ag/AgCl || Zn2+(x M) | Zn(s) develops a potential of -1.039 volts at 25oC, what is the Zn2+ concentration in the solution? (Answer in decimal format) Ag/AgCl reference electrode has E=0.199...
Please complete both!
1. When the following equation is balanced properly under acidic
conditions, what are the coefficients of the species shown?
____MnO2 + ____H3AsO3 --------_____Mn2+ +_____ H3AsO4
Water appears in the balanced equation as a_____ (reactant,
product, neither) with a coefficient of _____. (Enter 0 for
neither.) How many electrons are transferred in this
reaction?______
2.
A voltaic cell is constructed in which the anode is a Zn/Zn2+ half cell and the cathode is a Ag|Agt half cell. The...