At what $[O H^{-}]$ does the following half-reaction have a potential of $0 V$ when other species are at $1 M$ ?
$N O_{3}^{-} + H_{2} O + 2 e^{-} \to N O_{2}^{-} + 2 O H^{-}, E_{c e l l}^{\circ} = 0.01 V$ .

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Solution

$\begin{matrix} E_{c e l l} = E_{c e l l}^{^{\circ}} - \frac{0.0592}{2} log \frac{{[O H^{-}]}^{2} [N O_{2}^{-}]}{[N O_{3} O]} 0 \Rightarrow 0.01 - \frac{0.0592}{2} log \frac{{[O H^{-}]}^{2} \times 1}{1} \frac{0.0592}{2} \times 2 log [O H^{-}] = 0.01 log [O H^{-}] = 0.01 / 0.0592 = 0.169 [O H^{-}] = 10^{0.169} \Rightarrow 1.476 M \end{matrix}$

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Similar questions

Q. At what concentration of

[O H^{-}]

does the following half reaction has a potential of

0 V

when other species are at

1 M

N O_{3}^{-} + H_{2} O + 2 e^{-} \to N O_{2}^{-} + 2 O H^{-}; E^{0} c e l l = - 0.06 V

Q. The standard oxidation potential for the half cell

N O_{2}^{-} (g) + H_{2} O \to N O_{3}^{-} (aq) + 2 H^{+} (aq) + 2 e^{-}

- 0.78 V

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Calculate the reduction potential in

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assuming all other species at unit concentration. What will be the reduction potential in neutral medium?
(Given,

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Q. 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

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(i) Calculate the reduction potential in

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Assume all other species to be at unit concentration.

The standard reduction potential at $25^{o} C$ of the reaction $H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is $- 0 .8277 V$ . Calculate the equilibrium constant for the reaction $2 H_{2} O \to H_{3} O^{+} + O H^{-}$ at $25^{o} C$ .

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(10 \times 10^{- 7})

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At what [OH−] does the following half-reaction have a potential of 0V when other species are at 1 M?NO−3+H2O+2e−→NO−2+2OH−,E∘cell=0.01 V.

Ecell=E∘cell−0.05922log[OH−]2[NO−2][NO3O]0⇒0.01−0.05922log[OH−]2×110.05922×2log[OH−]=0.01log[OH−]=0.01/0.0592=0.169[OH−]=100.169⇒1.476M

At what $[O H^{-}]$ does the following half-reaction have a potential of $0 V$ when other species are at $1 M$ ?
$N O_{3}^{-} + H_{2} O + 2 e^{-} \to N O_{2}^{-} + 2 O H^{-}, E_{c e l l}^{\circ} = 0.01 V$ .