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:
<em>(1) </em>Cu²⁺(aq) +2e⁻ → Cu(s) E = 0.3419V
<em>(2) </em>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 = <em>0,62V</em>
2. Using Nernst equation:
Ecell = E°cell - 0,0591/n log Q
Where E° is <em>0,62V</em>, n are number of electrons <em>6e⁻ </em>And Q is: [Cu²⁺]³[NO]²/[H⁺]⁸[NO₃⁻]²
The Ecell is:
Ecell = 0,62V - 0,0591/6 log 3,92
<em>Ecell = 0,61V</em>
<em></em>
I hope it helps!