Al can be paired with other two to form cells , as the emf of
such cells are positive, which means spontaneous reaction.
1) Al/Fe cell
The cell reaction is
2Al + 3Fe+2 -----> 2 Al+3 + 3 Fe
For this cell
the emf = E0 of fe - E0 of Al= -0.44-(-1.66) = +1.22V
thus the cell is feasible.
2) Al/Sn cell
2Al + 3Sn+2 -----> 2 Al+3 + 3 Sn
For this cell
the emf = E0 of Sn - E0 of Al= -0.14-(-1.66) = +1.52V
thus the cell is feasible.
pls explain very detailed on how to do this question Question 6 Consider the following standard...
pls explain in great detail why
Consider the following standard reduction potentials Reduction Half-Reaction A1+ (aq) + 3e A1(s) Fe2+ (aq) + 2e-Fe(s) Sn2+ (aq) + 2e - Sn(s) E (volts) -1.66 0.44 0.14 The AIAP half-reaction can be paired with the other two to produce voltaic cells because Als* is a more powerful reducing agent OOO Al is a more powerful oxidizing agent o Fe and Sn are readily oxidized Al is a more powerful reducing agent AB+ is...
consider the following standard reduction potentials. Reduction Half-Reaction Eo (volts) Al3+(aq) + 3e− → Al(s) − 1.66 Fe2+(aq) + 2e− → Fe(s) − 0.44 Sn2+(aq) + 2e− → Sn(s) − 0.14 The Al/Al3+ half-reaction can be paired with the other two to produce voltaic cells because ________ A) Al is a more powerful oxidizing agent B) Fe and Sn are readily oxidized Al is a more powerful reducing agent C) Al3+ is a more powerful oxidizing agent D) Al3+...
I need help with questione 1-12 and discussion question 1 and
2. The previous pictures help determine the chart. Please Show Work
thank you so much
An oxidation half-reaction is characterized by electrons appearing on the product side. The oxidation of aluminum for instance would be represented thusly: Al(s) → Al3+ + 3e- (1) An reduction half-reaction is characterized by electrons appearing on the reactant side. The reduction of ferrous iron for instance would be represented thusly: Fe2+ + 2e...
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) +...
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...
Question 12 (10 points) Given the following standard reduction potentials, calculate Eºcell for the following reactions as written. Comment on the spontaneity of the reaction. El (Volts) +1.84 +1.36 +1.07 +0.45 +0.34 0.0 -0.14 -0.44 -0.76 Half-cell reaction CO3+ (aq) + el Co2+ (aq) Cl2 (9) + 2 = 2 CI (aq) Brą (aq) + 2 = 2 Brl(aq) SO2 (aq) + 4 H+ (aq) + 4e = S(s) + 2 H2) Cu2+ (aq) + 2 * = Cu (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)...
Consider the following standard reduction potentials in acid solution:E^o(V)Al3+ + 3e– ? Al(s) –1.66AgBr(s) + e– ? Ag(s) + Br– +0.07Sn4+ + 2e– ? Sn2+ +0.14Fe3+ + e– ? Fe2+ +0.77The strongest reducing agent among those shown above is
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...
Which of the following metal(s), if coated onto iron, would prevent the corrosion of iron? Standard Electrode Potentials at 25°C Reduction Half-Reaction Fe2+ (aq) +2 e Zn²+(aq) +2e Pb2+ (aq) +2e A13+ (aq) + 3e E° (V) -0.45 Fe(s) → Zn (s) → Pb (s) + Al(s) 오 오오 | Check all that apply. Zn O Pb Ο ΑΙ