Calculate the equilibrium constant, K, for the following reaction at 25˚C. 2 Fe3+(aq) + Cu(s) → 2 Fe2+(aq) + Cu2+(aq)           E° = 0.41 V

Nitesh

O2 (g) + 4H+(aq) + 4e- ⇌ 2 H2O (l) E10 = +1.23 V• Cu2+(aq) + 2 e- ⇌ Cu(s) E20 = +0.34 V• The equation for the oxidation of Cu(s) could be:• 2Cu(s) +O2(g) + 4H+ (aq) ⇌ 2 Cu2+ + 2 H2O (l) E30 = ? Find the E3 and E0 when the pH is raised to 7

The free energy change for the following reaction at 25 °C, when [Fe²*] = 1.18 M and [Cr³3+] = 8.62×10-3 M, is -186 kJ: 3FE2+(1.18 M) +2Cr(s)- →3F€(s) + 2C1³*(8.62×10-3 M) AG = -186 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?