Question

Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(aq) v (b) io3-(aq) + fe2+(aq) equilibrium reaction arrow fe3+(aq) + i2(aq)

Answer 1

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Step-by-step explanation:

We must look up the standard reduction potentials for the half-reactions.

ℰ°

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

ℰ°/V

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00

(b) Fe²⁺/IO₃⁻

ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

Step-by-step explanation:

its right trust

Answer 2

`\boxed{\text{(a) 0.00 V; (b) 0.424 V}}`

Step-by-step explanation:

We must look up the standard reduction potentials for the half-reactions.

ℰ°

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

ℰ°/V

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00

(b) Fe²⁺/IO₃⁻

ℰ°/V

Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424

The ℰ° values for the cells are
`\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}`