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Standard XII

Chemistry

Electrochemical Cells, Galvanic Cells

The concentra...

Question

The concentration of $K^{+}$ ion inside a biological cell is $20$ times higher than outside. The magnitude of potential difference between the two sides is : $[G i v e n : 2.303 \frac{R T}{F} = 59 m V]$

0 m V

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26 m V

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77 m V

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177 m V

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Solution

The correct option is C $77 m V$
According to Nernst equation,
$E = \frac{R T}{n F} l n \frac{C_{1}}{C_{2}}$
Here, the concentration of $K^{+}$ ion is $20$ times high.
So,
$\frac{C_{1}}{C_{2}} = 20$
$E = \frac{R T}{1 F} \times 2.303 l o g (20)$
$\Rightarrow E = 59 m V \times l o g (20)$
$= 59 \times 1.301$
$= 76.8 \approx 77 m V$

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The concentration of K+ ion inside a biological cell is 20 times higher than outside. The magnitude of potential difference between the two sides is : [Given:2.303RTF=59 mV]

The correct option is C 77 mVAccording to Nernst equation, E=RTnFlnC1C2 Here, the concentration of K+ ion is 20 times high. So, C1C2=20 E=RT1F×2.303 log (20) ⇒E=59 mV×log (20) =59×1.301 =76.8≈77 mV

The concentration of $K^{+}$ ion inside a biological cell is $20$ times higher than outside. The magnitude of potential difference between the two sides is : $[G i v e n : 2.303 \frac{R T}{F} = 59 m V]$