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Nernst Equation

If N O 3- ...

Question

If $N O_{3}^{-} \to N O_{2}$ (acidic medium); $E^{o} = 0.790 V$ and $N O_{3}^{-} \to N H_{2} O H$ (acidic medium); $E^{o} = 0.731 V$
At what pH the above two half reaction will have same E values?
Assume the concentrations of all the species to the unity and give an answer in volts in the nearest integer.

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Solution

The balanced half reactions are
$N O_{3}^{-} + 2 H^{+} + e^{-} \to N O_{2} + H_{2} O$
$N O_{3}^{-} + 7 H^{+} + 6 e^{-} \to N H_{2} O H + 2 H_{2} O$
$E_{N O_{3}^{-} | N O_{2}} = E_{N O_{3}^{-} | N H_{2} O H}$
$E_{N O_{3}^{-} | N O_{2}}^{0} - \frac{0.0592}{1} l o g \frac{1}{[H^{+}]^{2}} = E_{N O_{3}^{-} | N H_{2} O H}^{0} - \frac{0.0592}{6} l o g \frac{1}{[H^{+}]^{7}}$
$\frac{0.790 - 0.731}{0.0592} = l o g \frac{1}{[H^{+}]^{2}} - \frac{1}{6} l o g \frac{1}{[H^{+}]^{7}}$
$0.9966 = l o g \frac{1}{[H^{+}]^{2}} - \frac{1}{6} l o g \frac{1}{[H^{+}]^{7}}$
$[H^{+}] = 0.0632$
$p H = - l o g [H^{+}] = - l o g 0.0632 = 1.1992$

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Similar questions

{N O}_{3}^{-} \to {N O}_{2}

E^{o} = 0.79 V

{N O}_{3}^{-} \to {N H}_{2} O H

E^{o} = 0.731 V

E

H_{2} S + {N O}_{3}^{-} ⟶ N O_{2} + S_{8}

C r_{2} O_{7}^{- 2}

C r_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 C r^{+ 3} + 7 H_{2} O

[E^{\circ} {C r}_{2} O_{7}^{- 2} / C r^{+ 3} = 1.33 V]

C r_{2} O_{7}^{- 2}

C r_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \leftarrow 2 C r^{+ 3} + 7 H_{2} O

[E_{C r_{2} O_{7}^{- 2} / C r^{+ 3}}^{^{\circ}} = 1.33 V]

p H = 1

(E^{o})

A

E^{o} (V

I_{2} + 2 e^{-} \to 2 I^{-}; E^{o} = 0.54

C l_{2} + 2 e^{-} \to 2 C l^{-}; E^{o} = 1.36

M n^{3 +} + e^{-} \to M n^{2 +}; E^{o} = 1.50

F e^{3 +} + e^{-} \to F e^{2 +}; E^{o} = 0.77

O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O; E^{o} = 1.23

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