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Standard XII

Chemistry

Electrode Potential

At what conce...

Question

At what concentration of $[O H^{-}]$ does the following half reaction has a potential of $0 V$ when other species are at $1 M$ ?
$N O_{3}^{-} + H_{2} O + 2 e^{-} \to N O_{2}^{-} + 2 O H^{-}; E^{0} c e l l = - 0.06 V$

2.0 M

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1.0 M

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0.1 M

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0.01 M

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Solution

The correct option is A $2.0 M$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} log {[{O H}^{-}]}^{2}$
$0 = - 0.06 - \frac{0.591}{2} log {[{O H}^{-}]}^{2}$
$∴ [{O H}^{-}] = 10^{- 1} = 0.01 M$

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Similar questions

Q. 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

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

.
Calculate the reduction potential in

9

molar

H^{+}

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

log 9 = 0.95

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

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.

The standard reduction potential for the half cell :

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

0.78

V. What will be the reduction potential of the half cell in a neutral solution? Assume all the 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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At what concentration of [OH−] does the following half reaction has a potential of 0V when other species are at 1M?NO−3+H2O+2e−→NO−2+2OH−;E0cell=−0.06V

The correct option is A 2.0MEcell=E0cell−0.05912log[OH−]20=−0.06−0.5912log[OH−]2∴[OH−]=10−1=0.01M

At what concentration of $[O H^{-}]$ does the following half reaction has a potential of $0 V$ when other species are at $1 M$ ?
$N O_{3}^{-} + H_{2} O + 2 e^{-} \to N O_{2}^{-} + 2 O H^{-}; E^{0} c e l l = - 0.06 V$

The correct option is A $2.0 M$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} log {[{O H}^{-}]}^{2}$
$0 = - 0.06 - \frac{0.591}{2} log {[{O H}^{-}]}^{2}$
$∴ [{O H}^{-}] = 10^{- 1} = 0.01 M$