Problem: Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.N2H5+(aq) + 4Fe(CN)63-(aq)  →  N2(g) + 5H+(aq) + 4Fe(CN)64-(aq)

FREE Expert Solution

Step 1

N2(g) + 5H+ (aq) + 4e-  N2H5+ (aq)                              E° = - 0.23 V     E°  reduction  cathode

Fe(CN)64- (aq) + 6 H+ (aq) + 2 e-  Fe(s) + 6 HCN(aq)     E° = −1.16 V     E°  oxidation  anode

Step 2

cell = 0.93 V

Step 3

N2(g) + 5H+ (aq) + 4e-  N2H5+ (aq)

[Fe(s) + 6HCN(aq) ⇌ Fe(CN)64- (aq) + 6 H+ (aq) + 2 e-] x 2

N2(g) + 5H+ (aq) + 4e-  N2H5+ (aq)

2 Fe(s) + 12 HCN(aq) ⇌ 2 Fe(CN)64- (aq) + 12 H+ (aq) + 4 e-

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N2 (g) + 2 Fe (s) + 12 HCN(aq) ⇌ N2H5+ (aq)  + Fe(CN)64- (aq) + 7 H+ (aq)

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Problem Details

Using the standard reduction potentials listed in Appendix E in the textbook, calculate the equilibrium constant for each of the following reactions at 298 K.

N2H5+(aq) + 4Fe(CN)63-(aq)  →  N2(g) + 5H+(aq) + 4Fe(CN)64-(aq)