Given the listed standard electrode potentials, what is Eo for the cell ?4BiO+(aq)+3N2H5+(aq)⟶4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)N2(g)+5H+(aq)+4e−⟶N2H5+(aq), E∘=−0.23VBiO+(aq)+2H+(aq)+3e−⟶Bi(s)+H2O(l), E∘=+0.32V

N2(g)+5H+(aq)+4e−⟶N2H5+(aq), E∘=−0.23V…(1)BiO+(aq)+2H+(aq)+3e−⟶Bi(S)+H2O(l) E∘=0.32V…(2)[Reaction (1)] x -3 + [Reaction (ii) x 4, we get4BiO+(aq)+3N2H5+(aq)⟶4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)ΔG∘=−3−4FE∘+4−3FE∘−12FE∘=12F(−0.23)−12F(0.32)E∘=0.23+0.32=0.55

Given the listed standard electrode potentials, what is Eo for the cell ?4BiO+(aq)+3N2H5+(aq)⟶4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)N2(g)+5H+(aq)+4e−⟶N2H5+(aq), E∘=−0.23VBiO+(aq)+2H+(aq)+3e−⟶Bi(s)+H2O(l), E∘=+0.32V

N2(g)+5H+(aq)+4e−⟶N2H5+(aq), E∘=−0.23V…(1)BiO+(aq)+2H+(aq)+3e−⟶Bi(S)+H2O(l) E∘=0.32V…(2)[Reaction (1)] x -3 + [Reaction (ii) x 4, we get4BiO+(aq)+3N2H5+(aq)⟶4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)ΔG∘=−3−4FE∘+4−3FE∘−12FE∘=12F(−0.23)−12F(0.32)E∘=0.23+0.32=0.55

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Given the listed standard electrode potentials, what is E^o for the cell ?
$\begin{array}{l} 4 {BiO}^{+} (aq) + 3 N_{2} H_{5}^{+} (aq) ⟶ 4 Bi (s) + 3 N_{2} (g) + 4 H_{2} O (l) + 7 H^{+} (aq) \\ N_{2} (g) + 5 H^{+} (aq) + 4 e^{-} ⟶ N_{2} H_{5}^{+} (aq), E^{\circ} = - 0 .23 V \\ {BiO}^{+} (aq) + 2 H^{+} (aq) + 3 e^{-} ⟶ Bi (s) + H_{2} O (l), E^{\circ} = + 0 .32 V \end{array}$

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Detailed Solution

$\begin{array}{l} N_{2} (g) + 5 H^{+} (aq) + 4 e^{-} ⟶ N_{2} H_{5}^{+} (aq), E^{\circ} = - 0 .23 V \dots (1) \\ {BiO}^{+} (aq) + 2 H^{+} (aq) + 3 e^{-} ⟶ Bi (S) + H_{2} O (l) E^{\circ} = 0 .32 V \dots (2) \end{array}$
[Reaction (1)] x -3 + [Reaction (ii) x 4, we get
$\begin{array}{l} 4 {BiO}^{+} (aq) + 3 N_{2} H_{5}^{+} (aq) ⟶ 4 Bi (s) + 3 N_{2} (g) + 4 H_{2} O (l) + 7 H^{+} (aq) \\ \begin{array}{l} {ΔG}^{\circ} = - 3 (- 4 {FE}^{\circ}) + 4 - (3 {FE}^{\circ}) \\ - 12 {FE}^{\circ} = 12 F (- 0 .23) - 12 F (0 .32) \end{array} \\ E^{\circ} = 0 .23 + 0 .32 = 0 .55 \end{array}$