FInd E cell for reduction of NO3 - - NO by Cus- when - HNO3- - 1M-PNO - 10 - 3atm- Cu2 - - - 0-1MGiven- at 298 K RT-F 2-303 - 0-06volt-ENO3 - -NO - - - 0-96volt-ECu2 - -Cu - - - 0-34volt

The correct option is A ≈ 0.61 [ 2NO_3^ #160; + 3Cu → NO + Cu^2 +; ...

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Byju's Answer

Standard XII

Chemistry

Nernst Equation

FInd E cell ...

Question

FInd E $_{c e l l}$ for reduction of $N {O_{3}}^{-} \to N O$ by Cu(s), when $[H N O_{3}] = 1 M, P_{N O} = 10^{- 3} a t m, [C u^{2 +}] = 0.1 M$
(Given: at 298 K $\frac{R T}{F} (2.303) = 0.06 v o l t; E_{N O_{3}^{-} / N O}^{\circ} = + 0.96 v o l t; E_{C u^{2 +} / C u}^{\circ} = + 0.34 v o l t$ )

\approx 0.61

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\approx 0.71

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\approx 0.51

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\approx 0.81

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Solution

The correct option is A $\approx 0.61$
$\begin{matrix} 2 N {O_{3}}^{-} + 3 C u \to N O + C u^{2 +} n = 6 E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} log \frac{(C u^{2 +})}{(N O_{3}^{-})} = (0.96 - 0.34) - \frac{0.0591}{6} log (\frac{0.1}{1}) E_{c e l l} = 0.62 + 0.00985 E_{c e l l} \approx 0.62 \end{matrix}$

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Similar questions

Q. Copper reduces

N O_{3}^{-}

into

N O

and

N O_{2}

depending upon concentration of

H N O_{3}

in solution. Assuming

[C u^{2 +}] = 0.1 M

, and

P_{N O} = P_{N O_{2}} = 10^{- 3}

bar. At which concentration of

H N O_{3}

, thermodynamic tendency for reduction of

N O_{3}^{-}

into

N O

and

N O_{2}

by copper is same?
[Given:

E_{C u^{2 +} | C u}^{\circ} = + 0.34 V, E_{N O_{3}^{-} | N O}^{\circ} = + 0.96 V, E_{N O_{3}^{-} | N O_{2}}^{\circ} = + .79 V

Q. Copper reduces nitrate into

N O

and

N O_{2}

depending upon the concentration of

H N O_{3}

in solution (Assuming fixed

[C u^{2 +}]

and

P_{N O} = P_{N O_{2}}

). The

H N O_{3}

concentration at which the thermodynamic tendency for reduction of

N O_{3}^{-}

into

N O

and

N O_{2}

by copper is same is

10^{x}] M

. The value of

2 x

is
(Rounded-off to the nearest integer)
[Given :

E_{C u^{2 +} / c u}^{0} = 0.34 V, E_{N O_{3}^{-} / N O}^{0} = 0.96 V, E_{N O_{3}^{-} / N O_{2}}^{0} = 0.79 V

and at

\frac{R T}{F} (2.303) = 0.059

]

Q. For the cell:

T l_{(s)} | T l^{+} (0.001 M) | | C u^{2 +} (0.1 M) | C u_{(s)} :

E_{c e l l}

= 0.83 V at 298 K, The cell potential can be increased by

Below given are emf of three galvanic cells, represented by $E_{1}$ , $E_{2}$ and $E_{3}$ . Which of the following is true?

I. $Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u$

II. $Z n | Z n^{2 +} (0.1 M) | | C u^{2 +} (1 M) | C u$

III. $Z n | Z n^{2 +} (1 M) | | C u^{2 +} (0.1 M) | C u$

E

value for the cell reaction

C u | {C u}^{2 +} (0.001 M) | | {C u}^{2 +} (0.1 m) | C u

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FInd Ecell for reduction of NO3−→NO by Cu(s), when [HNO3]=1M,PNO=10−3atm,[Cu2+]=0.1M(Given: at 298 K RTF(2.303)=0.06volt;E∘NO−3/NO=+0.96volt;E∘Cu2+/Cu=+0.34volt)

The correct option is A ≈0.612NO3−+3Cu→NO+Cu2+n=6Ecell=Eocell−0.0591nlog(Cu2+)(NO−3)=(0.96−0.34)−0.05916log(0.11)Ecell=0.62+0.00985Ecell≈0.62

FInd E $_{c e l l}$ for reduction of $N {O_{3}}^{-} \to N O$ by Cu(s), when $[H N O_{3}] = 1 M, P_{N O} = 10^{- 3} a t m, [C u^{2 +}] = 0.1 M$
(Given: at 298 K $\frac{R T}{F} (2.303) = 0.06 v o l t; E_{N O_{3}^{-} / N O}^{\circ} = + 0.96 v o l t; E_{C u^{2 +} / C u}^{\circ} = + 0.34 v o l t$ )