Question

A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides. The following reaction describes the formation of NO, one of the products.3Cu(s) + 8H+(aq) + 2NO-3(aq) ------> 2NO(g) +3Cu2+ (aq) + 4H20(l)using the provided standard potentials, calculate E cell for this reactionCu2+(aq) +2e- ------> Cu(s) E=0.3419 V4H+(aq) + NO3-(aq) + 3e- -------> NO(g) + 2H20(l) E= 0.96 V__________VPart 2What is the value of Erxn at 208.00K when [H+]=0.100M, [NO3-]=0.02000m, [Cu2+]=0.0200M and the partial pressure of NO is 0.001400 atm?_________V

Answer 1

Answer:

1. E° = 0,62V

2. Ecell = 0,61V

Explanation:

1. It is possible to obtain the E° of a reaction by the sum of the E for the half-reactions, thus:

(1) Cu²⁺(aq) +2e⁻ → Cu(s) E = 0.3419V

(2) 4H⁺(aq) + NO₃⁻(aq) + 3e⁻ → NO(g) + 2H₂O(l) E = 0.96

As the E of (1) is less than E of (2), Cu(s) will be oxidated and the nitrogen will be reduced. That means:

Cu(s) → Cu²⁺(aq) +2e⁻ E = -0.3419V

4H⁺(aq) + NO₃⁻(aq) + 3e⁻ → NO(g) + 2H₂O(l) E = 0.96

The sum of the half-reactions is:

3x [Cu(s) → Cu²⁺(aq) +2e⁻] E = -0.3419V

2x [4H⁺(aq) + NO₃⁻(aq) + 3e⁻ → NO(g) + 2H₂O(l)] E = 0.96

3Cu(s) + 8H⁺(aq) + 6e⁻ + 2NO₃⁻(aq) → 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l) + 6e⁻

E° = -0,3419V + 0,96V = 0,62V

2. Using Nernst equation:

Ecell = E°cell - 0,0591/n log Q

Where E° is 0,62V, n are number of electrons 6e⁻ And Q is:  [Cu²⁺]³[NO]²/[H⁺]⁸[NO₃⁻]²

The Ecell is:

Ecell = 0,62V - 0,0591/6 log 3,92

Ecell = 0,61V

I hope it helps!