Given the listed standard electrode potentials, what is E0 for the cell :4BiO+(aq)+3N2H5+(aq)→4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)N2(g)+5H+(aq)+4e−→N2H5+(aq),E0=−0.23 VBiO+(aq)+2H+(aq)+3e−→Bi(s)+H2O(l), E0=+0.32 V

N2(g)+5H+(aq)+4e−→N2H5+(aq),E0=−0.23 V.......(1)BiO+(aq)+2H+(aq)+3e−→Bi(s)+H2O(l), E0=+0.32 V.......(2)[Reaction(1)]×−3+[Reaction(2)]×4,we get4BiO+(aq)+3N2H5+(aq)→4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)△G0=−3(−4FE0)+4(−3FE0)−12FE0=12F(−0.23)−12F(0.32)E0=0.23+0.32=0.55

Given the listed standard electrode potentials, what is E0 for the cell :4BiO+(aq)+3N2H5+(aq)→4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)N2(g)+5H+(aq)+4e−→N2H5+(aq),E0=−0.23 VBiO+(aq)+2H+(aq)+3e−→Bi(s)+H2O(l), E0=+0.32 V

N2(g)+5H+(aq)+4e−→N2H5+(aq),E0=−0.23 V.......(1)BiO+(aq)+2H+(aq)+3e−→Bi(s)+H2O(l), E0=+0.32 V.......(2)[Reaction(1)]×−3+[Reaction(2)]×4,we get4BiO+(aq)+3N2H5+(aq)→4Bi(s)+3N2(g)+4H2O(l)+7H+(aq)△G0=−3(−4FE0)+4(−3FE0)−12FE0=12F(−0.23)−12F(0.32)E0=0.23+0.32=0.55

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Given the listed standard electrode potentials, what is $E^{0}$ for the cell :
$\begin{array}{l} 4 B i O^{+} (a q) + 3 N_{2} {H_{5}}^{+} (a q) \to 4 B i (s) + 3 N_{2} (g) + 4 H_{2} O (l) + 7 H^{+} (a q) \\ N_{2} (g) + 5 H^{+} (a q) + 4 e^{-} \to N_{2} {H_{5}}^{+} (a q), E^{0} = - 0.23 V \\ B i O^{+} (a q) + 2 H^{+} (a q) + 3 e^{-} \to B i (s) + H_{2} O (l), E^{0} = + 0.32 V \end{array}$

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$+ 1.88$

$+ 0.55$

$+ 0.34$

$+ 0.09$

answer is A.

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Detailed Solution

$\begin{array}{l} N_{2} (g) + 5 H^{+} (a q) + 4 e^{-} \to N_{2} {H_{5}}^{+} (a q), E^{0} = - 0.23 V ....... (1) \\ B i O^{+} (a q) + 2 H^{+} (a q) + 3 e^{-} \to B i (s) + H_{2} O (l), E^{0} = + 0.32 V ....... (2) \\ [R e a c t i o n (1)] \times - 3 + [Re a c t i o n (2)] \times 4, w e g e t \\ 4 B i O^{+} (a q) + 3 N_{2} {H_{5}}^{+} (a q) \to 4 B i (s) + 3 N_{2} (g) + 4 H_{2} O (l) + 7 H^{+} (a q) \\ △ G^{0} = - 3 (- 4 F E^{0}) + 4 (- 3 F E^{0}) \\ - 12 F E^{0} = 12 F (- 0.23) - 12 F (0.32) \\ E^{0} = 0.23 + 0.32 = 0.55 \end{array}$