What is the value of the equilibrium constant for the cell reaction below at 25°C? Eºcell...
For the following cell, at 25 ° C, Cr (s) | Cr (NO3) 3 (0.200 M) || Pb (NO3) 2 (0.060 M) | Pb (s) the cell potential was 0.589 V. Calculate the equilibrium constant for the following reaction at this temperature : 2Cr (s) + 3Pb2 + (aq) --- 2 Cr3 + (aq) + 3Pb (s)
What is the value of the equilibrium constant for the cell reaction below at 25°C? Sn2+(aq) + Fe(s) ⇄ Sn(s) + Fe2+(aq)
Consider the following cell reaction: 2Cr(s) + 6H*(aq) - 2Cr3+ (aq) + 3H2(g); Eºcell = 0.74V Under standard-state conditions, what is Eº for the following half-reaction? C++ (aq) + 3e - Cr(s) Select one: a. -0.37 V Ob. 0.74 V Oc. -0.74 V Od 0.25V o e. 0.37 v
Question 2 (1 point) What is Eºcell for the cell reaction: 2Cr(s) + 3Sn4+(aq) --> 3Sn2+(aq) + 2Cr3+ (aq)? Given: Cr3+(aq) + 3e --> Cr(s); E° = -0.74 V Sn4+(aq) + 2e --> Sn2+(aq); E° = +0.15 V 0 +0.45 V O +0.89 V O 1.93 V 0-0.59 V 0 +0.59 V
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/
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: 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) +35n²+ (aq)
The free energy change for the following reaction at 25 °C, when [Pb2+) - 1.12 M and (Cr) - 4.80*10% M, is -369 kJ: 3Pb2+(1.12 M) + 2Cr(s) 3Pb(s) + 2Cr +(4.80X10-9M) AG --369 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?
For the balanced net redox reaction below, determine its standard cell potential (in V) at 25°C and whether the reaction is spontaneous at standard conditions. 3Ni2+(aq) + 2Cr(s) +3Ni(s) + 2Cr3+ (aq) -0.487, spontaneous 0.487 V, nonspontaneous -0.487, nonspontaneous 0.487 V, spontaneous
Calculate the equilibrium constant for each of the reactions at 25 ∘C. a. 2Cr3+(aq)+3Sn(s)→2Cr(s)+3Sn2+(aq) b. O2(g)+2H2O(l)+2Sn2+(aq)→4OH−(aq)+2Sn4+(aq) c. 2Cr3+(aq)+3Ni(s)→2Cr(s)+3Ni2+(aq)
6) Calculate the equilibrium constant K at 25°C for the following reaction for tant K at 25°C for the following reaction for the standard cell potential (7 points) (nFEⓇ - RT In K, F=96485 C/mol.R=8.31 J/molk) Pb2+ (aq) + Fe(s) 5 Pb(8) + Fe²(aq) 7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s) | Fe*(aq) (1.1 M) || Cu?" (aq) (0.50 M) Cu() Ecell - Eºcell = 0.0592/n logQ