For the above cell-E-cell- 0E-cell- - 0- triangle G 0E-cell- - 0- triangle G-circ- - 0

In the above electrolytic cell we have-Ecell - -x2212-0-0591 log-oxi-red-Ecell - -x2212-0-0591 log0-051Ecell- 0-13599 V-xA0-x394-G- -x2212-nFEcell-x394-G is negative

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Standard XII

Chemistry

Gibb's Energy and Nernst Equation

Question

For the above cell.
$E_{c e l l} < 0; △ G > 0$
$E_{c e l l} > 0; △ G < 0$
$E_{c e l l} < 0; △ G^{\circ} > 0$
$E_{c e l l} > 0; △ G^{\circ} < 0$

E_{c e l l} < 0; △ G > 0

E_{c e l l} > 0; △ G^{\circ} < 0

E_{c e l l} > 0; △ G < 0

E_{c e l l} < 0; △ G^{\circ} > 0

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Solution

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In the above electrolytic cell we have,
$E_{c e l l}$ = $- 0.0591$ $l o g$ $\frac{(o x i)}{(r e d)}$

$E_{c e l l}$ = $- 0.0591$ $l o g$ $\frac{0.05}{1}$

$E_{c e l l}$ = 0.13599 V

$Δ$ G= $- n F E_{c e l l}$
$Δ$ G is negative

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The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is:
$M (s) | M^{+}$ (aq;0.05 molar)|| $M^{+}$ (aq;1 molar)|M(s)
For the above electrolytic cell the magnitude of the cell potential $| E_{c e l l} | = 70 m V$ .

For the above cell

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Assertion :Current stops flowing when

E_{c e l l} = 0

. Reason: Equilibrium of the cell reaction is attained.

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During the working of a galvanic cell and with the passage of time:

1) Spontaneity of the cell reaction decreases, $E_{c e l l}$ decreases.

2) Reaction quotient Q decreases, $E_{c e l l}$ increases.

3) Reaction quotient Q increases, $E_{c e l l}$ decreases.

4) At equilibrium, $Q = K_{e q}, E_{c e l l} = 0$ .

Identify the correct statement(s).

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Assertion :

E_{c e l l}^{\circ} = 0

for a chloride ion concentration cell. Reason: For this concentration cell:

E_{c e l l} = \frac{R T}{n F} l n \frac{[C l^{-}]_{L H S}}{[C l^{-}]_{R H S}}

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Assertion :A reaction is spontaneous if

E_{c e l l} = + v e

. Reason: For

E_{c e l l} = + v e, △ G

is always

- v e

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For the above cell.Ecell<0;△G>0Ecell>0;△G<0Ecell<0;△G∘>0Ecell>0;△G∘<0

In the above electrolytic cell we have,Ecell = −0.0591 log(oxi)(red)Ecell = −0.0591 log0.051Ecell= 0.13599 V ΔG= −nFEcellΔG is negative

For the above cell.
$E_{c e l l} < 0; △ G > 0$
$E_{c e l l} > 0; △ G < 0$
$E_{c e l l} < 0; △ G^{\circ} > 0$
$E_{c e l l} > 0; △ G^{\circ} < 0$

In the above electrolytic cell we have,
$E_{c e l l}$ = $- 0.0591$ $l o g$ $\frac{(o x i)}{(r e d)}$

$E_{c e l l}$ = $- 0.0591$ $l o g$ $\frac{0.05}{1}$

$E_{c e l l}$ = 0.13599 V

$Δ$ G= $- n F E_{c e l l}$
$Δ$ G is negative