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Chemistry
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15 Nov 2019

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. [ Fe3+(aq) + 3 e → Fe (s) E° = -0.04 V ] 3 I2(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 I⁻(aq)

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Related questions

Half Reaction

Cr3+ (aq) + 3e- → Cr (s) E° = -0.74 V

Fe2+ (aq) + 2e- → Fe (s) E° = -0.440 V

Fe3+ (aq) + e- → Fe2+ (s) E° = +0.771 V

Sn4+ (aq) + 2e- → Sn2+ (aq) E° = +0.154 V

The standard cell potential (E°cell) for the voltaic cell based on the reaction below is __________ V.

2Cr (s) + 3Fe2+ (aq) → 3Fe (s) + 2Cr3+ (aq)

Reaction E°, V

Mg2+ (aq) + 2e– → Mg (s) -2.356

Fe2+ (aq) + 2e– → Fe (s) -0.440

Pb2+ (aq) + 2e– → Pb (s) -0.125

2 H+ (aq) + 2e– → H2 (g) 0.0

Cu2+ (aq) + 2e– → Cu (s) +0.337

I2 (s) + 2e– → 2 I– (s) +0.535

Fe3+ (aq) + 2e– → Fe2+ (aq) +0.771

Ag+ (aq) + e– → Ag (s) +0.800

O2 (g) + 4 H+ (aq) + 2e– → 2 H2O +1.229

Cl2 (g) + 2e– → 2 Cl– (aq) +1.358

F2 (g) + 2e– → 2 F– (aq) +2.866

Calculate the value of Keq at 25°C for the reaction:

F2 (g) + 2 I– ↔ 2 F– (aq) + I2 (g)

A) 6.0 × 1078

B) 1.4 × 10–79

C) 2.33

D) 2.6 × 1039

E) 5.3 × 1039

Please Explain!!

Calculate the equilibrium constant for each of the reactions at 25∘C.
Standard Electrode Potentials at 25 ∘C
Reduction Half-Reaction E∘(V)
Fe3+(aq)+3e− →Fe(s) -0.036
Pb2+(aq)+2e− →Pb(s) -0.13
Ni2+(aq)+2e− →Ni(s) -0.23
O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40
Cl2(g)+2e− →2Cl− 1.36
I2(s)+2e− →2I− 0.54

Part A

2Fe3+(aq)+3Pb(s)→2Fe(s)+3Pb2+(aq)

Express your answer using two significant figures.
K =

Incorrect; Try Again; 4 attempts remaining

Part B

O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq)

Express your answer using two significant figures.
K =

Part C

Cl2(g)+2I−(aq)→2Cl−(aq)+I2(s)

Express your answer using two significant figures.
K =

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