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What is the value of AG° at 25.00 °C for the voltaic cell based on the...
Please use the following to answer the below: A fictional galvanic (voltaic) cell consists of an electrode composed of a metal, M, in a 1.0 mol/L M2+ ion solution and a second electrode composed of an unreactive platinum metal (Pt) in a 1.0 mol/L X– ion solution, connected by a salt bridge and an external wire at 25.0 ⁰C. Consider the two entries below from a fictional table of standard reduction potentials. Reduction Equation Standard reduction potentials (E⁰reduction) M2+ (aq)...
A voltaic cell is set up with one beaker containing 1.0 M Zn(NO 3) 2 and a zinc electrode, and another beaker containing 1.0 M Ni(NO 3) 2 and a nickel electrode. Given the following standard reduction potentials, answer the 3 questions below: Eº Zn2+(aq) + 2e → Zn(s) -0.76 V Ni2+(aq) + 2e → Ni(s) -0.23V Part a. Write out the half-cell reaction that occurs at the anode of the voltaic cell. Part b. In which direction do...
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
Given the following standard reduction potentials choose the cell which will work as a voltaic cell. All cells below are written according to the usual cell diagram convention. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V 2H+(aq) + 2e → H2(g) E° = 0.00 V Sn2+ (aq) + 2e → Sn(s) E° = -0.14 V Ni2+(aq) + 2e → Ni(s) E° = -0.26 V Cd2+(aq) + 2e → → Cd(s) E° = -0.40 V Sn(s) | Sn2+(aq) || Ni2+(aq)...
A voltaic cell represented by the following cell diagram has Ecell= 1.409 V . Zn(s)|Zn2+(1.00M)||Ag+(saturated Ag2X)|Ag(s) What must be the Ksp of Ag2X? Use the following standard electrode potentials: Zn2+(aq) + 2e- →Zn(s) E° = -0.764 V Ag+(aq) + e- →Ag(s) E° = 0.799 V To enter your answer, multiply your Ksp by 1x109 then enter it to 2 decimal places.
38. The following redox half reactions are combined in a voltaic cell. Which reaction occurs at the cathode and what is the Eceu? Fe2+(aq) + 2e → Fe(s) E°=-0.44 V Cu²+(aq) + 2e → Cu(s) E°= 0.34 V a) b) c) d) Cu2+(aq) + 2e → Cu(s), Ecell = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecel = 0.78 V Fe2+(aq) + 2e → Fe(s), Ecell =-0.10 V Cu²+(aq) + 2e → Cu(s), Ecel = 0.10 V Cu²+ (aq) +...
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...
A voltaic cell is based on the following two half-reactions: Cd2 (ag) +2e-> Cd (s) Sn2(aq)+ 2e Sn (s) Calculate the standard cell potential. Use the date from the attached table.SRP2.docx Oa 0.13 Ob 042 Oc.027 Od-0.27
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)...
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...