Answer:
[tex]\boxed{\text{(a) 0.00 V; (b) 0.424 V}}[/tex]
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 [tex]\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}[/tex]
