HomeworkLib
Question

What is the effect on the emf of the cell shown in Figure 20.9, which has...

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?

  1. The pressure of the H2H2 gas is increased in the cathode half-cell.

  2. Zinc nitrate is added to the anode half-cell.

  3. Sodium hydroxide is added to the cathode half-cell, decreasing [H+].[H+].

  4. The surface area of the anode is doubled.

Can someone please explain why b is emf decreases? Why would these ions have an effect on the voltaic system?

science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

Answer

a) Decrease

b) Decrease

c) Decrease

d) No effect

Explanation

Zn(s) + 2H+ (aq) -------> Zn2+(aq) + H2(g)

Reaction quotient, Q = [Zn2+ ​​​​​​]× PH2/ [H+]

number of electron transfer, n = 2

Nernst equation at 25℃ is

Ecell = E°cell - (0.0592V/n)logQ

From this equation we know that , cell potential depends on Q

If concentration Zn2+ is icreased , cell potential would decresae

Zn2+ concentration is increased by adding zinc nitrate to anode half -cell

So, cell emf decreased by adding zinc nitrate solution to anode half cell.

0 0
Add a comment
Know the answer?
Add Answer to:
What is the effect on the emf of the cell shown in Figure 20.9, which has...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • For a voltaic cell based on the reaction below, which statement is correct? Zn(s)+2H+(aq)→Zn2+(aq)+H2(g) Zn2+(aq) is...

    For a voltaic cell based on the reaction below, which statement is correct? Zn(s)+2H+(aq)→Zn2+(aq)+H2(g) Zn2+(aq) is oxidized at the anode. H+(aq) is reduced at the cathode. Zn2+(aq) is reduced at the anode. H2(g) is the oxidizing agent.

  • 1a) In the following spontaneous voltaic cell, which of the following statements are TRUE? Select as...

    1a) In the following spontaneous voltaic cell, which of the following statements are TRUE? Select as many answers as are correct however points will be deducted for incorrect guesses. Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s) Select one or more: Zn/Zn2+ is the anode Eºcell is positive H+/H2 is the cathode Zn/Zn2+ is the cathode H+/H2 is the anode Eºcell is negative

  • In a copper-zinc voltaic cell, one half-cell consists of a ZnZn electrode inserted in a solution...

    In a copper-zinc voltaic cell, one half-cell consists of a ZnZn electrode inserted in a solution of zinc sulfate and the other half-cell consists of a CuCu electrode inserted in a copper sulfate solution. These two half-cells are separated by a salt bridge. At the zinc electrode (anode), ZnZn metal undergoes oxidation by losing two electrons and enters the solution as Zn2+Zn2+ ions. The oxidation half-cell reaction that takes place at the anode is Zn(s)→Zn2+(aq)+2e−Zn(s)→Zn2+(aq)+2e− The CuCu ions undergo reduction...

  • Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2:...

    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);...

  • Part A Calculate the pH of the cathode compartment solution if the cell emf at 298...

    Part A Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.640 V when [Zn240.27 M and PH = 0.85 atm. Express your answer using two decimal places. V ΑΣφ pH =$ Figure 1 of 1 Previous Answers Request Answer Submit Switch 8.76 X Incorrect; Try Again; 2 attempts remaining Voltmeter Zn anode NO Na - H2(g) Cathode compartment (standard hydrogen electrode) NO3 Zn2 Provide Feedback Next Anode compartment NO3...

  • 1. Calculate the pH of the cathode compartment solution if the cell emf at 298 K...

    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)

  • The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is...

    The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V: Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2 (g) With PH2 = 1.0 atm and [Zn2+] = 1.0 M, the cell potential is 0.42 V. The concentration of H+ in the cathode compartment is ________ M. please explain the math Thank you

  • Consider a voltaic cell al 25 degree Celsius in which the reaction is: Zn (s) +...

    Consider a voltaic cell al 25 degree Celsius in which the reaction is: Zn (s) + 2H+ (aq) → Zn2+ (aq) + H2(g) It is found that the voltage (think Ecell) is +0.560 V when [Zn2+1 -0.85 M and PH2 = = 0.988 atm. What is the pH in the H2-H half-cell?

  • In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of...

    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] = 0.042 M [H]- 19 M partial pressure of H2 =0.37 atm. Calculate Ecell at 298 K (enter to 3 decimal places). Zn2 (aq) + 2e +2H (aq) + 2e1 Eo-0.76 V E 0.00 V Zn(s) H2(g)

  • In a test of a new reference electrode, a chemist constructs a voltaic cell consisting of...

    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...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT