9.15 The electrode isn't working at standard conditions so we'll use Nernst Law to obtain the electrode potentials at both concentrations and then obtain the change:
that for the electrode half-reaction
becomes:
so:
9.16 Solid manganese oxidizes, so this is the anodic reaction:
Chlorine gas gets reduced so this the cathodic reaction
The standard cell potential is defined by:
we have to use the standard potential of the cell and of the Cl2/Cl- couple (cathodic reaction) to obtain the standard potential of the anodic reaction (Mn2+/Mn couple), so:
9.16:
help please 9.15 Consider a hydrogen electrode in HBr(aq) Estimate the change in the electro de...
solve all please and only one per question just for me to get the idea 9.18 Use the standard potentials of the electrodes to calculate the standard potentials of the cells in Exercise 9.17. 9.19 Devise cells in which the following are the reactions. In each case state the value for v to use in the Nernst equation. (a) Fe(s)+PbS04(aq) FeSO4(aq) + Pb(s) (b) Hg2Cl2(s) H2(g) 2 HCl(aq) +2 Hg(I) + 9.20 Use the standard potentials of the electrodes to...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
Standard potentials are measured against the standard hydrogen electrode (SHE). Because it is not always convenient to use a S.Н.Е., often other reference electrodes are used. The saturated calomel electrode (SCE.) is one commonly used reference electrode, with a reduction potential of +0.242 V versus the S.H.E. Using a table of standard reductions, determine what the standard reduction potential of each reduction half-reaction would be versus the S.C.E Cl2 (g) + 2e-→ 2 Cl-(aq) E1.36 Fe3 + (aq) + 3...
use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 C. (The equation is balanced.) 3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq) Express your answer to two significant figures and include the appropriate units. em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...
looking for help for question 1 please Experiment 23 Electrochemical Cells Postlaboratory Questions Lell potentials are state functions so the cell potential for Zn (s)[Zn- (aq) Fe (aq me added to the cell potential for Fe+ (aq) Fe (s) Cu²+ (aq)Cu (s) should equal the cell potential of Zn (s)IZn2+ (aq)||Cu2+ (aq) Cu (s). a. Does this hold using the theoretical cell potentials? b. Does this hold using the experimentally determined cell potentials? c. Why might the two differ? 2....
need help for half cell potentials pls calculate step by step (NOTE - Remember that the positive electrode is attached to the red wire and the negative electrode is attached to the black wire.) Electrode Systems Used Anode (oxidation) Negative Cathode (reduction) Positive Measured Potential (V) Positive (Ecu) Copper and silver Cu() - Cu2+ + 2e" Ag+ +1e Ag) 0.432 V Zinc + Silver Zn cs + 2n**+ Zé dat + leº nAg (s) 1.484 V Copper & Zinc Zn(s)...
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
An aqueous Snl, solution is electrolyzed under 1 bar pressure using platinum electrodes. (a) Write the half-reactions predicted to occur at the anode and cathode, based on the standard cell potentials given below. Standard Reduction Potentials (Volts) at 25°C 12(s) +2 — 21'(aq) 0.535 O2(g) + 4 H30* (aq) + 4 —6H20(1) 1.229 2 H20(1) + 2 € H2(g) + 2 OH'(aq) -0.828 Sn²+(aq) +2 € + Sn(s) -0.140 Half-reaction at anode: Half-reaction at cathode: (b) What is the expected...
Calculate the standard free-energy change at 25 ∘C for the following reaction: Mg2+(aq) + Zn(s) → Mg(s) + Zn2+(aq) Express your answer to three significant figures and in units of kJ/mol. Consider constructing a voltaic cell with one compartment containing a Zn(s) electrode immersed in a Zn2+ aqueous solution and the other compartment containing an Al(s) electrode immersed in an Al3+ aqueous solution. What is the spontaneous reaction in this cell? Group of answer choices Zn + Al3+ → Al...
If the potential of a hydrogen electrode based on the half-reaction 2 H^+ (aq) + 2 e^- rightarrow H_2 (g) is 0.000 V at pH = 0.00, what is the potential of the same electrode at pH = 6.75? Changing pH is changing the concentration of H^+. When the concentration is changed, the Nernst equation allows us to calculate the new cell potential.