The single electrode potential of metal $E$ of $0.1$ M solution of M $^{+} / M$ ions [ $E$ $^{0}$ $= - 2.36 V$ ] is nearly :

+ 2.41 V

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- 2.41 V

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4.82 V

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+ 4.82 V

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Solution

The correct option is B $- 2.41 V$
$E_{c e l l} = E_{c e l l}^{\circ} + \frac{0.059}{n} l o g [M^{+}]$

$= - 2.36 + 0.059 l o g (0.1)$

$E_{c e l l} = - 2.41 V$

Option B is correct.

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Similar questions

Q. The reduction electrode potential

E

0.1

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(E_{k p}^{o} = - 2.36 V)

is :

Q. The reduction electrode potential, E of

0.1

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The single electrode potential of $0.1 M$ solution of $M^{+} / M$ ion is:

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A zinc electrode is dipped in a 0.1 M solution at $25^{0} C$ . Assuming that salt is dissociated to 20% at this dilution. Calculate the electrode potential. $E {^{0}}_{Z n^{2 +} / Z n} = - 0.76 V$ .

The name of equation showing relation between electrode potential (E)

standard electrode potential and concentration of ions in solution is

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The single electrode potential of metal E of 0.1 M solution of M+/Mions [E0=−2.36V] is nearly :

The correct option is B −2.41VEcell=E∘cell+0.059nlog[M+]=−2.36+0.059log(0.1)Ecell=−2.41VOption B is correct.

The single electrode potential of metal $E$ of $0.1$ M solution of M $^{+} / M$ ions [ $E$ $^{0}$ $= - 2.36 V$ ] is nearly :

The correct option is B $- 2.41 V$
$E_{c e l l} = E_{c e l l}^{\circ} + \frac{0.059}{n} l o g [M^{+}]$

$= - 2.36 + 0.059 l o g (0.1)$

$E_{c e l l} = - 2.41 V$

Option B is correct.