What is the value of the equilibrium constant for the cell reaction below at 25°C? Sn2+(aq) + Fe(s) ⇄ Sn(s) + Fe2+(aq)
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What is the value of the equilibrium constant for the cell reaction below at 25°C? Sn2+(aq)...
Use the data in the table below to calculate the equilibrium constant at 25°C for the reaction: Cl2(g) + Sn(aq) + Sn2+ (aq) + 2C1- (aq) Standard Reduction Potentials at 25°C Sn(aq) + 2e + Sn2+ (aq) E° = -0.14 V Cl2 (g) + 2e + 2C1- (aq) E° = 1.36 V Express your answer to two significant figures. O ALQ O a ?
What is the value of the equilibrium constant for the cell reaction below at 25°C? Eºcell = 0.61 VE 3Pb(s) + 2Cr3+(aq)2Cr(s) + 3Pb2+(aq)
Use standard reduction potentials to calculate the equilibrium constant for the reaction: Fe(a 2Fe(a) Fe(s)+2Fe (aq) Hint: Carry at least Equilibrium constant: than zero. AO for this reaction would be 9 more group a Submit Answer Retry Entire Group ing Use standard reduction potentials to calculate the equilibrium constant for the reaction: Sn2+(aq) + Fe(s)-→ Sn(s) + Fe2+(aq) Hint: Carry at least S significant figures during intermediate calculations to avoid round off error when taking the antilogarithnm. Equilibrium constant AG°...
Question 30 (6 points) What is the equilibrium constant (K) at 25°C for the following cell reaction? Sn(s) + Pb2+ (aq) → Sn2+ (aq) + Pb(s); Eºcell = 0.014 V (Please submit your detail solution work to the dropbox, which can be found in the assignments.) A/
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...
Question 1) What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Pb(s) and Sn2+(aq) to give Sn(s) and Pb2+(aq) Use the reduction potential values for Sn2+(aq) of -0.14 V and for Pb2+(aq) of -0.13 V Question 2) What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Cr(s) and Cu2+(aq) to give Cu(s) and Cr3+(aq) Use the reduction potentials for Cr3+(aq) is -0.74 V and...
The free energy change for the following reaction at 25°C, when [Sn2+] = 1.18 M and [Fe2+] = 4.88 10-M.is -71.5 kJ: Sn2+(1.18 M) + Fe(s)— Sn(s) + Fe2+(4.88 10-3M) AG = -71.5 kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction?
Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe (s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn** (aq) +2Cr(s) → 2Cr** (aq) +3Sn²+ (aq) 3. What is the cell potential, Ecell, for the reaction above if [Sn] = 1.00 M, [Cr3+1 = 0.0200 M and [Sn2+] = 0.0100 M?
The free energy change for the following reaction at 25 °C, when [Sn2+] = 1.17 M and [Fe2+] = 6.64×10-3 M, is -70.7 kJ: Sn2+(1.17 M) + Fe(s) Sn(s) + Fe2+(6.64×10-3 M) ΔG = -70.7 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?
the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) fone 2 pts) a. Use the standard half-cell potentials listed below to calculate the standard cell potential (Eºcell) for 3 Sn(s) + 2 Fe* (aq) - 3 Sn2+ (aq) + 2 Fe(s) Sn 2(aq) + 2 e -Sn (s) E = -0.14 V Fe3+ (aq) + 3 e Fe(s) E° = -0.036 V A) -0.176 V B)-0.104 V C) +0.104 V D) +0.176 V b. Write...