Consider the reaction Fe + Sn2+(1 × 10–3 M) ? Fe2+(1.0 M) + Sn. Calculate the voltage theoretically generated by the cell. _______V Fe2++2e?Fe E0=?0.44V Sn2++2e?Sn E0=?0.14V Selected Answer: 0.30
Can someone tell me why this answer is wrong and the correct one?
Consider the reaction Fe + Sn2+(1 × 10–3 M) ? Fe2+(1.0 M) + Sn. Calculate the...
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?
Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...
Selective Oxidation The standard reduction potential for the half-reaction Sn4+ + 2e - Sn2+ is +0.15 V. Consider data from the table of standard reduction potentials for common half-reactions, in your text. For a galvanic cell under standard conditions, which of the following cathodic half reactions would produce, at the anode, a spontaneous oxidation of Sn to Sn2+ but not Sn2+ to Sn4+. 2H+ + 2e - H2 Fe3+ + 3e + Fe Sn2+ + 2e Fe2+ + 2e →...
ans A C B how ? 14. Consider the titration of 25.00 mL of 0.0500 M Sn2+ with 0.100 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+, using Pt and saturated calomel electrodes SCE l Sn4+, Sn2+, Fe3+, Fe3 | Pt(s) The titration reaction: 2Fe3 + Sn2+Snt+ 2Fe2+ The two half-reactions for the indicator electrode: Fe3+e Sn 2e Sn2+ Indicate the two Nernst equations for the cell voltage. E 0.732 V E 0.139 V Fe2+ IFe2 log...
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?
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?
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°...
Consider the reaction: Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 49 ∘C , where [Fe2+]= 3.80 M and [Mg2+]= 0.310 M . Part E What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 49 ∘C when [Fe2+]= 3.80 M and [Mg2+]= 0.310 M . Express your answer to three significant figures and include the appropriate units.
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 49 ∘C , where [Fe2+]= 3.30 M and [Mg2+]= 0.210 M . What is the value for the reaction quotient, Q, for the cell? Express your answer numerically.
The free energy change for the following reaction at 25°C, when [Fe2+1 = 8.61*10' M and (Sn2t - 1.17 M. is 70.1 kJ Fe2+(8.61x10 3 M) + Sn(s) Fe(s) + Sn2+(1.17 M) AG = 70.1 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? Submit Answer Retry Entire Group 3 more group attempts remaining The free energy change for the following reaction...