Quinhydrone electrode is sometimes used to find the pH of a solution- It is based on the following s -2 H - aq -2 e -sQuinone hydroquinoneIf standard electrode potential is 0-70 V- If in a particular solution- the electrode potential is found to be 0-58 V- the pH of the solution isA- 2B- 6C- 8D- 4

The correct option is A 2For the given electrode reaction E=E^∘ 0.0591/2log [ 1/[H^+]^2 ] =E^∘ + 0.0591 log[H^+] =E^∘ 0.0591 pH ∴ 0.58=0.70 0.0591 pH o ...

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Byju's Answer

Standard XII

Chemistry

Faraday's First Law

Quinhydrone e...

Question

Quinhydrone electrode is sometimes used to find the pH of a solution. It is based on the following

If standard electrode potential is 0.70 V. If in a particular solution, the electrode potential is found to be 0.58V, the pH of the solution is

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Solution

The correct option is A 2
For the given electrode reaction
$E = E^{\circ} - \frac{0.0591}{2} l o g [\frac{1}{[H^{+}]^{2}}]$
$= E^{\circ} + 0.0591 l o g [H^{+}]$
$= E^{\circ} - 0.0591 p H$
$∴ 0.58 = 0.70 - 0.0591 p H$
or, $p H = \frac{0.12}{0.059} = 2$

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Quinhydrone electrode is sometimes used to find the pH of a solution. It is based on the following If standard electrode potential is 0.70 V. If in a particular solution, the electrode potential is found to be 0.58V, the pH of the solution is

The correct option is A 2For the given electrode reaction E=E∘−0.05912log[1[H+]2] =E∘+0.0591 log[H+] =E∘−0.0591 pH ∴0.58=0.70−0.0591 pH or, pH=0.120.059=2

Quinhydrone electrode is sometimes used to find the pH of a solution. It is based on the following

If standard electrode potential is 0.70 V. If in a particular solution, the electrode potential is found to be 0.58V, the pH of the solution is

The correct option is A 2
For the given electrode reaction
$E = E^{\circ} - \frac{0.0591}{2} l o g [\frac{1}{[H^{+}]^{2}}]$
$= E^{\circ} + 0.0591 l o g [H^{+}]$
$= E^{\circ} - 0.0591 p H$
$∴ 0.58 = 0.70 - 0.0591 p H$
or, $p H = \frac{0.12}{0.059} = 2$