Calculate the equilibrium constant for this reaction at 25º C,
Al (s) + Fe3+ (aq) —> Al3+ (aq) + Fe (s)
given the following standard reduction potentials for the two half-reactions.
Fe3+ (aq) + 3 e- —> Fe (s) Eº = 0.77 V
Al3+ (aq) + 3 e- —> Al (s) Eº = -1.66 V
Calculate the equilibrium constant for this reaction at 25º C, Al (s) + Fe3+ (aq) —>...
Calculate the equilibrium constant, ?, for the reaction shown at 25 °C. Fe3+(aq)+B(s)+6H2O(l)⟶Fe(s)+H3BO3(s)+3H3O+(aq) The balanced reduction half‑reactions for the equation and their respective standard reduction potential values (?∘) are Fe3+(aq)+3e−⟶Fe(s) ?∘=−0.04V H3BO3(s)+3H3O+(aq)+3e−⟶B(s)+6H2O(l) ?∘=−0.8698 V K=
1A) the equilibrium constant of reaction
Sr(s)+Mg24(ac)---><---
Sr24(ac)+Mg(s)
is 2.69X1012 at 25º C calculate the Eº of the cell
formed by the half cells Sr / Sr2 + and Mg / Mg2
+
R= 8.314 J/k -mol
F= 96500 J/V- mol
1B)
Calculate the standard emf of a cell using the Ag / Ag1 + and Al
/ Al3 + half-cell reactions.
answers
A)0.86 V
B)2.46 V
C)-1.86 V
D)-0.86 V
E cell AGⓇ =-nFEcell Ecell = F Ink AG°...
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...
Using the reduction potentials given, calculate the equilibrium
constant, K, at 20 degrees C for the reaction
Using the reduction potentials given, calculate the equilibrium constant, K, at 25°C for the reaction, 33 3+ Ag (aa) t Fe(a)Ag) Fe (aq) +0.77 V +0.80 V A Ag+(aq) + e- ← a. 1.66 b. 6.4 c. 3.2 d. 6.1 x 10-4 e. 1.6 x 104 Rank the following compounds according to increasing solubility in water. K” is a less than sign) 34...
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+...
A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc. of Fe2+= 0.402 M, Fe3+= 0.073 M and Al3+= 0.239 M. Use the reduction potentials for Al3+ is -1.66 V and for Fe3+ is 0.77 V. How do you decide which is oxi, red.
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...
The equilibrium constant, K, for a redox reaction is related to the standard potential, Eº, by the equation In K = nFE° RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e), R (the gas constant) is equal to 8.314 J/(mol · K), and T is the Kelvin temperature. Standard reduction potentials Reduction half-reaction E° (V) Ag+ (aq) + e +Ag(s) 0.80 Cu²+ (aq) + 2e + Cu(s) 0.34...
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)...
Cell Potential and Equilibrium Standard reduction potentials The equilibrium constant, K, for a redox reaction is related to the standard cell potential, Ecel, by the equation Reduction half-reaction (V) Ag+ (aq) + e-→Ag(s) Cu2+ (aq) + 2e-→Cu(s) 0.34 Sn (a) 4e-Sn(s 0.15 2H' (aq) + 2e-→H2 (g) Ni2+ (aq) + 2e-→Ni(s)-0.26 Fe2+ (aq) + 2e-→Fe(s)-0.45 Zn2+ (aq) + 2e-→Zn(s)-0.76 Al3+ (aq) +3e-→Al(s) -1.66 Mg2+ (aq) + 2e-→Mg(s) -2.37 0.80 n FEcell where n is the number of moles of electrons...