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)

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

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.