Acopper 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 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

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!

answer: they produce more slow than unicellular organisms.