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Standard XII

Chemistry

Nernst Equation

A graph is pl...

Question

A graph is plotted between $E_{c e l l}$ and
$l o g_{10} [\frac{Z n^{2 +}}{C u^{2 +}}]$ The curve was linear with intercept
on $E_{c e l l}$ axis equal to 1.10 V. Calculate $E_{c e l l}$ for
$Z n ∣ ∣_{0.1 M}^{Z n^{2 +}} ∣ ∣ ∣ ∣_{0.01 M}^{C u^{2 +}} ∣ ∣ C u$ :

1.07 V

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1.129 V

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0.059 V

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0.0295 V

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Solution

The correct option is A 1.07 V
The intercept is equal to $E_{c e l l}^{0}$ .
Hence, $E_{c e l l}^{0} = 1.10 V$ .
The Nernst equation for the emf of the cell is $E_{c e l l}^{0} = E_{c e l l}^{0} - \frac{0.0592}{n} l o g \frac{⌊ {Z n}^{2 +} ⌋}{⌊ {C u}^{2 +} ⌋}$ .
Substitute values in the above equation.
$E_{c e l l}^{0} = 1.10 V - \frac{0.0592}{2} l o g \frac{⌊ 0.1 ⌋}{⌊ 0.001 ⌋} = 1.07 V .$

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Similar questions

Q. EMF of Daniell cell was found using different concentrations of

{Z n}^{2 +}

ion and

{C u}^{2 +}

ion. A graph was then plotted between

E_{c e l l}

and

log \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]}

. The plot was found to be linear with intercept in

E_{c e l l}

axis equal to

1.10 V

E_{c e l l}

for

Z n / {Z n}^{2 +}

(0.1 M) ∥ {C u}^{2 +} (0.01 M)

will be:

Q. Plots of

E_{c e l l}

with

l o g_{10} Q

for the reaction,

Z n (s) + C u^{2 +} (a q .) ⇌ C u (s) + Z n^{2 +} (a q .)

gives a straight line with an intercept 1.10. If

E_{c e l l} = 1.1591 V

, then value of Q, i.e.

\frac{[Z n^{2 +}]}{[C u^{2 +}]}

is:

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

Reaction quotient is

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

. Variation of

E_{c e l l}

with log Q is of the type with OA = 1.10

V . E_{c e l l}

will be

1.1591 V

when:

Z n (s) + C u^{2 +} (a q .) ⇌ C u (s) + Z n^{2 +} (a q .)

Reaction quotient

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

, variation

E_{c e l l}

with

Q

is of the type with

O A = 1.10 v o l t . E_{c e l l} = 1.1591 v o l t

when:

$Z n + C u {^{2 +}}_{(a q)} ⇌ C u + Z n {^{2 +}}_{(a q)}$

Reaction quotient, $Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

Variation of $E_{c e l l}$ with log Q is of the type as shown in the graph with $O A = 1.10 V . E_{c e l l}$ of the reaction will be 1.1591 V when:

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A graph is plotted between Ecell andlog10[Zn2+Cu2+] The curve was linear with intercepton Ecell axis equal to 1.10 V. Calculate Ecell forZn∣∣Zn2+0.1M∣∣∣∣Cu2+0.01M∣∣Cu:

The correct option is A 1.07 VThe intercept is equal to E0cell.Hence, E0cell=1.10V .The Nernst equation for the emf of the cell is E0cell=E0cell−0.0592nlog⌊Zn2+⌋⌊Cu2+⌋.Substitute values in the above equation.E0cell=1.10V−0.05922log⌊0.1⌋⌊0.001⌋=1.07V.

A graph is plotted between $E_{c e l l}$ and
$l o g_{10} [\frac{Z n^{2 +}}{C u^{2 +}}]$ The curve was linear with intercept
on $E_{c e l l}$ axis equal to 1.10 V. Calculate $E_{c e l l}$ for
$Z n ∣ ∣_{0.1 M}^{Z n^{2 +}} ∣ ∣ ∣ ∣_{0.01 M}^{C u^{2 +}} ∣ ∣ C u$ :