*Label the parts of the voltaic cells below=(given the Evalues at bottom)cathode, anode, salt bridge, wire, voltmeter.* Draw on arrow with o colorful pen to show the flow of electrons.Then use the space provided to calculate the Ę and write it in the voltmeter.Au+3(aq)Cu+2+ Cu(s)+ Au (s)(aq)6. Calculate the E。A-...Au(s)1.BVDAu¹³ (ovCLP 2 (ovCu(s)асcu1.anode3.salt bridge5. voltmeter2cathode4. wireAut+2(oq)(aq) + 3e-+ 2e-Cu (s)SRP = 0.34 VAu(s) SRP = 1.498 V

Answered: Image /qna-images/answer/43170e0b-250c-4a08-8683-d67f39a64ef0.jpg

Answered: *Label the parts of the voltaic cells…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Part B The half-reactions that occur in ordinary alkaline batteries can be written as Cathode : MnO2 (s) + H20 (1) +e+MnO (OH) (s) + OH (aq) Anode : Zn (s) + 20H (ag) →Zn(OH), (s) + 2e In 1999, researchers in Israel reported a new type of alkaline battery, called a "super-iron" battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO ion (from K2FEO4) to solid Fe(OH), at the cathode. Write a balanced equation for the cathode half-reaction in a super-iron battery. The half-reaction occurs in a basic environment. O Fe(OH): (s) + 50H-(ag)→FE0- (ag) + 4H2O(1) + 3e- O FeO; (aq) + 4H2O(1) + 3e →Fe(OH);(8) + 50H (aq) O FeO (aq) + H2O(1) + 3EFE(OH) (s) + OH-(aq) O FeO (ag) + 2H2O(1)+ 3eFe(OH);(s) + 20H (aq) Submit Request Answer Part C A super-iron battery should last longer than an ordinary alkaline battery of the same size and weight because its cathode can provide more charge per unit mass. Quantitatively compare the number of…
Electrolysis of molten NaClNaCl is done in a Downs cell operating at 7.0 volts (V) and 5.0×104 AA . How much Na(s)Na(s) and Cl2Cl2 can be produced in eight hours in such a cell? Enter the mass of NaNa and mass of Cl2Cl2 in grams, separated by a comma.
13) The average C−C bond dissociation energy (D) is 350 kJ/mol and the average C=C bond dissociation energy is 728 kJ/mol. Based on these values, which is stronger: a σ or a π bond?
When attempting to create a voltaic cell that will produce electrical work, if a Sn2*/ Sn**couple is used as an anode what cathodic couple would produce the most electrical work? Half-Reaction E (V Clg)+ 2e 20 (ag) Cr0, (ag) + 14H" (ag) + 6e 2C "(ag) + 7H,O) MnO,() + 4H"(ag) + 2e Mn"(ag) + 2H,O) NO, ag) + 4H'(ag) + 3e NOg) + 2HOU) Ag'ag) Fe"(ag) +e Cu"(ag) + 2e = Cu(s) AgC + So, (ag) + 4H"(ag) + 2e SO,(g) + 2H,O) Cu*(ag) +e= Cu tag) Sn" (ag) + 2e Sn"(ag) +1.36 +1.33 +1.23 +0.96 Agis) +0.80 +0.77 +0.34 +0.22 +0.20 = Agir) + CT (ag +0.15 +0.13 0.00 Ph"(ag) + 2e= Phis) Sn"(ag) + 2e Sels) 2e Ni) -0.13 -0.14 -0.25 -031 -0.40 -0.44 PhSOA) + H"(ag) + 2e P(s) + HSO, (ag) Ca"(ag) + 2e Ca(s) Fels) C"(ag) + 3e Cs) Za"(ag) + 2e Zals) -0.74 -0.76 O Cu2*/Cu* O Ni2+/Ni O Cu2+/Cu O Fe3+/Fe2+
Out of the following statements about galvanic cells that you use in a laboratory determine which one is false 1 . you need a salt bride to determine electroneutrality 2. If you switch the voltmeter leads between the cathode and anode it will change the sign of the observed charge 3. In a Cu^2+ / Cu(s) (aq) half cell, Cu(s) would be the electrode 4. Eocell=Eoanode+Eocathode
An electrochemical cell is based on these two half - reactions Ox : Pb(s) -> Pb ^ 2 (aq, 0.20 mol L^ -1 )+2 e^ -; E anode ^ 0 =-0.13(V) Red: MnO4 ^ - (aq,1.55 mol L^ -1 )+4 H^ + (aq,2.0 mol L^ -1 )+3 e^ - MnO^2(s) + 2H_{2}*O(l) E 1.68 (V) Compute the cell potential at 25 ∘C.
6) Write the anode and cathode half reactions and calculate E°cell for the following redox reaction; Sn(s) + 2Cu²“(aq) → Sn²*(aq) + 2Cu*(aq) Ered(Cư-/Cu*)= 0.15 V E red(Sr*/Sn) = -0.14 V
Predict the potential for the following voltaic cell at 25 C:Cu(s)|CuSO4(1.29 M)||AgNO3(0.7 M)|Ag(s)
a) Balance the following redox reaction in basic solution: H2O2(aq) + ClO2(aq) → ClO2-(aq) + O2(g) b) Sketch the voltaic cell for the redox reaction, labeling the direction for electron flow, the anode and cathode , and the species present in each half-cell. Show all the work please!
4
If you watched the "Breaking Bad" video, the two components of the battery that he was manufacturing were mercuric oxide (HgO) and zinc. Here are the half reactions for this battery. Consider the following two half reactions: HgO (s) + H₂O (1) + 2e → Hg(s) + 2OH(aq) Eºr red = 0.0977 V Zn(OH)42 (aq) + 2e → Zn(s) + 4OH(aq) Ered = -1.19 V Take these two half reactions and make a full redox reaction. What is the expression for the reaction quotient (Q) when you're done? Q Q = [Zn(OH)² ][Hg] [OH-]²[Zn][HgO] [OH-] [Zn] [HO] [Zn(OH)2](Hg) Q = [OH-]2 Zn(OH)2 Zn(OH)2 [OH-]2
4) A voltaic cell is constructed using the reaction a) Sketch the voltaic cell. Zn (s) + Ni¹2 (aq) → Zn¹2 (aq) + Ni (s) b) Which half-reaction occurs at the anode? c) Which half-reaction occurs at the cathode? d) Calculate Eºcell.

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