The ERP for the reaction- 2Cl-aq- - Cl2g - 2e- at Pt electrode is-

The correct options are A E_RP = E^o_RP + 0.0592 log [Cl^ ]^2P_Cl_2 B E_RP = E^o_RP 0.0592 log P_Cl_2[Cl^ ]^2 ...

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Standard XII

Chemistry

Equivalent, Molar Conductivity and Cell Constant

The ERP for...

Question

The $E_{R P}$ for the reaction, $2 C l^{-} (a q .) \to C l_{2} (g) + 2 e^{-}$ ; at Pt electrode is:

E_{R P} = E_{R P}^{o} + \frac{0.059}{2} l o g \frac{[C l^{-}]^{2}}{P_{C l_{2}}}

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E_{R P} = E_{R P}^{o} - \frac{0.059}{2} l o g \frac{P_{C l_{2}}}{[C l^{-}]^{2}}

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E_{R P} = E_{R P}^{o} + \frac{0.059}{1} l o g \frac{[C l^{-}]^{2}}{P_{C l_{2}}}

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E_{R P} = E_{R P}^{o} - 0.059 l o g \frac{[P_{C l_{2}}]^{1 / 2}}{[C l^{-}]}

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Solution

The correct options are
A $E_{R P} = E_{R P}^{o} + \frac{0.059}{2} l o g \frac{[C l^{-}]^{2}}{P_{C l_{2}}}$
B $E_{R P} = E_{R P}^{o} - \frac{0.059}{2} l o g \frac{P_{C l_{2}}}{[C l^{-}]^{2}}$

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Similar questions

Q. Assertion :The standard reduction potential of

M^{n +} / M

electrode increases with increase in activity of

M^{n +}

ion. Reason: The standard reduction potential is given by:

E_{R P} = E_{R P}^{o} + \frac{0.059}{n} l o g [M^{n +}]

Q. ERP stands for _______________.

Q. Consider the cell

Z n / Z n^{2 +} (a q .) | | C l^{-} (a q) / C l_{2} (g) | 1 p t (s)

Given

[Z n^{+ 2}] = 4 M; [C l^{-}] = 0.5 M, (P_{C l_{2}}) = 0.1 b a r

E_{Z n / Z n^{2 +}}^{\circ} = 0.76 V; E_{C l_{2} / C l^{-}}^{\circ} = 1.32 V

\frac{2.303 R T}{F} = 0.06

The electrical work done during the reaction at 298 K:
$2 H g_{(l)} + C l_{2 (g)} \to H g_{2} C l_{2 (s)}, i s G i v e n t h a t E_{\frac{C l_{2}}{C l^{-}}}^{\circ} = 1.36 V; E_{\frac{H g_{2} C l_{2}}{H g, C l^{-}}}^{\circ} = 0.27 V; P_{C l_{2}} = 1 a t m,$

Q. Determine the standard reduction potential for the half reaction.

{C l}_{2} + {2 e}^{-} \to {2 C l}^{-}

Given:

{P t}^{2 +} + {2 C l}^{-} \to P t + {C l}_{2}, E_{c e l l} = - 0.15 V

{P t}^{2 +} + {2 e}^{-} \to P t, E = 1.20 V

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The ERP for the reaction, 2Cl−(aq.)→Cl2(g)+2e−; at Pt electrode is:

The correct options are A ERP=EoRP+0.0592log[Cl−]2PCl2 B ERP=EoRP−0.0592logPCl2[Cl−]2

The $E_{R P}$ for the reaction, $2 C l^{-} (a q .) \to C l_{2} (g) + 2 e^{-}$ ; at Pt electrode is:

The correct options are
A $E_{R P} = E_{R P}^{o} + \frac{0.059}{2} l o g \frac{[C l^{-}]^{2}}{P_{C l_{2}}}$
B $E_{R P} = E_{R P}^{o} - \frac{0.059}{2} l o g \frac{P_{C l_{2}}}{[C l^{-}]^{2}}$