Question 12 (10 points) Given the following standard reduction potentials, calculate Eºcell for the following reactions...
Using the standard reduction potentials given below, choose the reaction than can only be achieved through electrolysis. Cu2+(aq) + 2e → Cu(s) E° = 0.34 V Pb2+(aq) + 2e + Pb(s) E° = -0.13 V Fe2+(aq) + 2e Fe(s) E° = -0.44 V Zn2+(aq) + 2e + Zn(s) E° = -0.77 V Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq) o Fe2+(aq) + Zn(s) → Fe(s) + Zn2+(aq) Pb2+(aq) + Fe(s) → Pb(s) + Fe2+(aq) Cu2+(aq) + Fe(s) → Cu(s) +...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
please help with these, the standard reduction potentials are given in the last two pictures A standard galvanic cell is constructed in which a Cu2+ Cu half cell acts as the cathode. Which of the following statements are correct? Hint: Refer to a table of standard reduction potentials. (Choose all that apply.) The cathode reaction is Cu -> Cu2+ + 2e Fe2+Fe could be the other standard half cell. Hg2+Hg could be the other standard half cell. In the external...
Using the standard reduction potentials given below, decide which of the following reactions will occur spontaneously as written. Fe3+(aq) + e + Fe2+(aq) E° = 0.77 V Sn4+ (aq) + 2e → Sn2+(aq) E° = 0.13V Zn2+(aq) + 2e → Zn(s) E° = -0.77 V Lit(aq) + e + Li(s) E° = -3.05 V 2Li+ (aq) + Sn2+(aq) → 2Li(s) + Sn4+(aq) Sn2+(aq) + Zn2+(aq) → Sn4+(aq) + Zn(s) Lit(aq) + Fe2+(aq) - Li(s) + Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) →...
If a box is not needed, leave it blank Use the table Standard Reduction Potentials' located in the 'Tables', to predict if a reaction will occur between Ni metal and C12(g), when the two are brought in contact via half-cells in a voltaic cell. If a reaction will occur, write a balanced net ionic equation for the reaction, assuming that the productions are in aqueous solution. If no reaction will occur, leave all boxes blank. Submit Answer Retry Entire Group...
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)...
Using the standard reduction potentials listed, calculate the equilibrium constant for each of the following reactions at 298 K. A) Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) Express your answer using two significant figures. B) Co(s)+2H+(aq)→Co2+(aq)+H2(g) Express your answer using two significant figures. C) 10Br−(aq)+2MnO−4(aq)+16H+(aq)→2Mn2+(aq)+8H2O(l)+5Br2(l) Express your answer using two significant figure. E°(V) -0.83 +0.88 +1.78 +0.79 Half-Reaction E°(V) Half-Reaction Ag+ (aq) + - Ag(s) +0.80 2 H20(1) + 2 e — H2(8) + 2 OH+ (aq) AgBr(s) + - Ag(s) + Br" (aq) +0.10 HO2...
2. Calculate the standard cell potential for each of the following redox reactions, and then predict whether each will occur spontaneously as written. a. Sr(s) + Fe2+(aq) → Sr2+ (aq) + Fe(s) b. 2Cr(s) + 3Cd2+ (aq) → 2Cr3+ (aq) + 3Cd(s) 3. Calculate the standard cell potential, Eºcell, for each of the voltaic cells in Part II of the experiment. a. Zn(s) | Zn2+ (aq, 1.0 M) || Cu2+ (aq, 1.0 M) Cu(s) b. Zn(s) | Zn2+(aq, 1.0 M)...
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 • Which of these half-cells would combine with the SHE to give the largest possible Eºcell value? • Fe3+/Fe, Ni2+/Ni, Cu2+/Cu = = = Zn2+ (aq) + 2e Cr+ (aq) + 3e Fe2+ (aq) + 2 Cd + (aq) + 2e Ni”* (aq) + 2e Sn²+ (aq) + 2e Pb2+ (aq) + 2 Fe+ (aq) + 3 2H(aq) + 2 Sn+ (aq) + 2e Cu²+ (aq) + e Cu²(aq) + 2e Zn(s) Cr(s) Fe(s) Cd(s) Ni(s) Sn(s) Pb(s) Fe(s)...