materials question electrochemistry
table has associated voltages
compute the voltage at 25c of an electrochemical cell
consisting of pure copper immersed in 9x10^-2M solution of Cu^2+
ions and pure nickel in a 0.35 M solution of Ni^2+ ions.
half reactions are:
Cu-> Cu^2+ + 2e- @+0.345V electrode potential
Ni-> Ni^2+ + 2e- @ -2.250V
this is all the information I was given
Thanku, please upvote.
materials question electrochemistry table has associated voltages compute the voltage at 25c of an electrochemical cell...
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...
19 20 Question 16 Half-cell Potentials: Half Reaction: E' value + 0.80 V +0.77 V Agte → AS Fe3+ + + Fe2+ Cu2+ 2e → Cu Pb2+ + 2e → Pb +0.34 V -0.13 V Ni2+ + 2e → NI -0.25 V - 0.40 V Cd2+ +2e → ca Fe2+ + 2e → Fe Zn2+ + 2e → Zn -0.44 V - 0.76 V A13+ +3 → AI - 1.66 V Consider an electrochemical cell constructed from the following half...
need help with the rest of the table EXPERIMENT 10 DETERMINATION OF THE ELECTROCHEMICAL SERIES PURPOSE To determine the standard cell potential values of several electrochemical coll INTRODUCTION The basis for an electrochemical cell is an oxidation reduction Corredor be divided into two half reactions reaction. This reaction can Oxidation half reaction Gloss of electrons) takes place at the anode, which is the positive electrode that the anions migrate to Chence the name anode) Reduction half reaction (gain of electrons)...
Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...
An electrochemical cell is made of a left-hand compartment consisting of nickel wire in contact with a 1.0 M Ni(ClO), solution and a right-hand compartment of copper wire immersed in 1.0 M Cu(Cl) 1. What is the overall spontaneous cell reaction for this electrochemical cell? What is the standard cell potential for this cell? Show your work for full credit. 2. 3. What is the direction of electron flow (from which electrode to which)? Which substance is oxidized? 4. If...
5. The nickel-cadmium battery, despite issues with the toxicity of Cd, is still used in some rechargeable batteries because it has very consistent voltage over long discharge times. a) Using only half-cell reactions given in Table 16.1, write the overall reaction and give the highest AV° value possible for a NiCd battery. b) The actual half-cell reactions used in a Nicd battery are: Ca(OH)2(8) + 2e → Cds + 2OH(aq) (Vº = -0.86V) NiO(OH)) + H2O) +e Ni(OH)2(s) + OH(aq)...
need answers for all questions Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) // Cu2+ (aq) / Cu (+), the correct half reactions are 0 (a) Cr3+ (a) Cr3+ + 3e → Crat (-) electrode; Cu+ 2e → Cu2+ at (+) electrode (b) Cr - 3e → Cr3+ at (-) electrode; Cu2+ + 2e → 2 Cu at (+) electrode Occ) cr (c) Cr + 3e → Cr3+ at (-) electrode; Cu2+ - 2e → Cu...
1. Which metal can reduce Ni2+ ions but not Fe2+ ions? Cd²⁺,Cr,Sn²⁺,Sn,Cd 2.Which species can oxidize Sn but not Fe? Ni,Ni²⁺,Pb²⁺,Cd,Pb 1.68 1.51 1.50 1.09 1.00 0.96 0.95 FAO) + 2 H,0,1a) + 2H(aq) + 2e PO4 + 4H"(aq) + S0,00) + 2€ Mno. (aq) + 4H(aq) + 3e Mno,aq) + 8H(aq) + 5e Au?+ (aq) + 3€ PbO (8) + 4H(aq) + 2 Cl@) + 2 Cr,0,?" (aq) + 14 H*(aq) + 6e 0,(@) + 4H(aq) + 4e MnO...
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)...
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...