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Byju's Answer

Standard XII

Chemistry

pH the Power of H

The reduction...

Question

The reduction potential of hydrogen electrode when placed in a buffer solution is found to be 0.413V. The part of the buffer is:

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Solution

The correct option is C 10
Given: Reduction Potential of the electrode is 0.413 V.
$E = E_{\circ} - \frac{R T}{n F} l o g [H^{+}]$
$E = E_{\circ} - \frac{R T}{n F} p H$
$0.413 = 0 + 0.0591 p H$
$p H = 0.413 / 0.0591 = 6.988$ $\approx$ $7$
Hence, $p H$ is $7$ .

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Q. The reduction potential of hydrogen electrode when placed in a buffer solution is found to be - 0.413V. The pH of the buffer is:

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A hydrogen electrode placed in a buffer solution of $C H_{3} C O O N a$ and $C H_{3} C O O H$ in the ratios $x : y$ and $y : x$ has electrode potential values $E_{1}$ volts and $E_{2}$ volts respectively at 25 $^{0}$ C . The $p K_{a}$ value of acetic acid is :

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The reduction potential of hydrogen electrode when placed in a buffer solution is found to be 0.413V. The part of the buffer is:

The correct option is C 10Given: Reduction Potential of the electrode is 0.413 V.E=E∘−RTnFlog[H+]E=E∘−RTnFpH0.413=0+0.0591pHpH=0.413/0.0591=6.988 ≈ 7Hence, pH is 7.

The correct option is C 10
Given: Reduction Potential of the electrode is 0.413 V.
$E = E_{\circ} - \frac{R T}{n F} l o g [H^{+}]$
$E = E_{\circ} - \frac{R T}{n F} p H$
$0.413 = 0 + 0.0591 p H$
$p H = 0.413 / 0.0591 = 6.988$ $\approx$ $7$
Hence, $p H$ is $7$ .