The standard reduction potential for the half-cell having reaction,
$N O_{3}^{-} (a q .) + 2 H^{+} (a q .) + e^{-} \to N O_{2} (g) + H_{2} O$ is $0.78 v o l t$ .
(i) Calculate the reduction potential in $8 M H^{+}$ .
(ii) What will be the reduction potential of the half-cell in a neutral solution?
Assume all other species to be at unit concentration.

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Solution

(i) Applying the formula,
$E_{r e d} = E_{r e d}^{\circ} + \frac{0.0591}{n} log [H^{+}]^{2}$
$= 0.78 + \frac{0.0591}{2} log 8^{2}$
$= 0.78 + 0.0591 \times 3 \times 0.3010$
$= 0.833 v o l t$
(ii) $E_{r e d} = 0.78 + \frac{0.0591}{2} log (10^{- 7})^{2}$
[For neutral solution $[H^{+}] = 10^{- 7} M$ ]
$= 0.78 - 0.0591 \times 7$
$= 0.367 v o l t$ .

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Similar questions

N O_{3}^{-} (a q) + 2 H^{+} (a q) + e^{-} ⟶ N O_{2} (g) + H_{2} O

0.78

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The reduction potential in 8 M $H^{\oplus}$ is: (in V)

Assume all the other species to be at unit concentration. ( write your answer to nearest integer)

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