For the above cell-E-cell- 0E-cell- - 0 - Delta G 0 - Delta G-o 0

M-s-M-aq-0-05M-M-aq-1M-M-s-Ecell-70mV means may be -ve or -veAccording to Nernst equation-E-Eo-x2212-0-0591logQ-nEcell-EOPLHS-ERPRHS-EoOP-x2212-0-0591log-M-LHS-EoRP-0-0591log-M-RHS-0-059log-M-RHS-M-LHS-0-059log10-05-0-059-xD7-1-30-0-076V-76mV-70mVEcell-ve and-x394-G-x2212-nFE-x2212-veSo for this reaction-Ecell-gt-0-x394-G-lt-0

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Standard X

Chemistry

Electrolysis and Electrolytes

Question

For the above cell:
$E_{c e l l} < 0 : Δ G > 0$
$E_{c e l l} > 0 : Δ G < 0$
$E_{c e l l} > 0 : Δ G^{o} < 0$
$E_{c e l l} < 0 : Δ G^{o} > 0$

E_{c e l l} > 0 : Δ G^{o} < 0

E_{c e l l} < 0 : Δ G > 0

E_{c e l l} > 0 : Δ G < 0

E_{c e l l} < 0 : Δ G^{o} > 0

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Solution

Verified by Toppr

$M (s) | M_{a q}^{+} (0.05 M) | | M_{a q}^{+} (1 M) | M (s)$

$| E_{c e l l} | = 70 m V$ means may be +ve or -ve

According to Nernst equation:

$E = E^{o} - 0.0591 l o g Q / n$

$E_{c e l l} = E_{O P_{L H S}} + E_{R P_{R H S}}$

$= E_{O P}^{o} - \frac{0.059}{1} l o g [M^{+}]_{L H S} + E_{R P}^{o} + \frac{0.059}{1} l o g [M^{+}]_{R H S}$

$= 0.059 l o g \frac{[M^{+}]_{R H S}}{[M^{+}]_{L H S}} = 0.059 l o g \frac{1}{0.05}$

$= 0.059 \times 1.30 = 0.076 V = 76 m V = 70 m V$

$E_{c e l l} = + v e$ and $Δ G = - n F E = - v e$

So for this reaction,

$E_{c e l l} > 0; Δ G < 0$

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For the above cell:Ecell<0:ΔG>0Ecell>0:ΔG<0Ecell>0:ΔGo<0Ecell<0:ΔGo>0

For the above cell:
$E_{c e l l} < 0 : Δ G > 0$
$E_{c e l l} > 0 : Δ G < 0$
$E_{c e l l} > 0 : Δ G^{o} < 0$
$E_{c e l l} < 0 : Δ G^{o} > 0$