By Sunil Bhardwaj

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Quinhydrone is bronze coloured needle shaped crystals, It is sparingly soluble in water. It is equimolar mixture of quinone and hydroquinone. It is the example of redox electrode.

Construction: With the help of this electrode we can find out the pH of the solution. Some definite volume say 25ml of the test solution (whose pH is to be determined) is taken in a beaker. Then saturate this acid solution with the quinhydrone. Now, insert the Pt wire for electrical contact. This assembly is known as Quinhydrone Electrode.

Representation: $$ Pt, { H }_{ 2 }Q, Q | Acid \ Soln, pH = ? Or Pt, { C }_{ 6 }{ H }_{ 4 }{ \left( OH \right) }_{ 2 }, { C }_{ 6 }{ H }_{ 4 }{ O }_{ 2 } | Acid \ Soln, pH = ? $$

Working: This electrode is known as indicator electrode. It is connected with reference electrode i.e. SCE, we get the cell $$ SCE \parallel Acid \ Soln, pH = ? | Q, { H }_{ 2 }Q, Pt $$ This cell is to be connected to potentiometer and we find out the emf of the cell. $$ { E }_{ cell } = \underset { Quinhydrone }{ { E }_{ Red } } - \underset { SCE }{ { E }_{ Red } } \qquad ....(2) $$ for reduction of quinhydrone, $$ { Q }_{ (s) } + 2{ H }^{ + } + 2{ e }^{ - } \longrightarrow { H }_{ 2 }{ Q }_{ (s) } $$ On applying the law of mass action: $$ K = \frac { { a }_{ { H }_{ 2 }{ Q }_{ (s) } } }{ { a }_{ { Q }_{ (s) } } \times { a }_{ { H }^{ + } }^{ 2 } } = \frac { 1 }{ { a }_{ { H }^{ + } }^{ 2 } } $$ As it is equimolar mixture Applying nernst equation, $$ \underset { Quinhydrone }{ { E }_{ Red } } = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \frac { RT }{ 2F } \ln { \frac { 1 }{ { a }_{ { H }^{ + } }^{ 2 } } } $$ $$ = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \frac { RT }{ 2F } \ln { { \left( { a }_{ { H }^{ + } } \right) }^{ -2 } } $$ $$ = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \frac { RT }{ 2F } \times \left( -2 \ln { \left( { a }_{ { H }^{ + } } \right) } \right) $$ $$ = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \frac { RT }{ F } pH $$ $$ = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - 0.059 pH $$ Putting this in equation (2) $$ { E }_{ cell } = \left( \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - 0.059 pH \right) - \underset { SCE }{ { E }_{ Red } } $$ $$0.059 pH = \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \underset { SCE }{ { E }_{ Red } } - { E }_{ cell } $$ $$\therefore pH = \frac { \underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } - \underset { SCE }{ { E }_{ Red } } - { E }_{ cell } }{ 0.059 } $$ But \(\underset { Quinhydrone }{ { E }_{ Red }^{ 0 } } = 0.7V \) and \( \underset { SCE }{ { E }_{ Red } } = 0.242V\) $$\therefore pH = \frac { 0.7 - 0.242 - { E }_{ cell } }{ 0.059 } $$ $$ = \frac { 0.458 - { E }_{ cell } }{ 0.059 } $$ Thus pH of the solution can be calculated using the Quinhydrone electrode.

Advantages:

(1) The electrode is very easy to setup.

(2) It attains the equilibrium very quickly.

(3) The temperature coefficient value i.e. \(\frac { dE }{ dT }\) is small.

(4) It can be used for non-aqueous media also.

Limitations:

(1) In a solution state the activity of quinone and hydroquinone must be unity.

(2) See that addition of quinhydrone to the experimental solution does not bring out any change in the experimental solution.

(3) As it is redox electrode, therefore the presence of oxidising or reducing substance must be excluded from experimental solution.

(4) The electrode is working very successfully up to the pH of 8.5 to 9. If the pH is more than 9, then \({ a }_{ Q } > { a }_{ { H }_{ 2 }Q }\). Hence it does not give correct value of pH.