The cell reaction involving the quinhydrone electrode is given above:
What will be the electrode potential at $p H = 3$ ?

1.48 V

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1.20 V

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1.10 V

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1.30 V

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Solution

The correct option is A $1.48 V$
Given Data : 1) Standard Electrode Potential = $E^{\circ}$ = $1.30$ $v o l t$
2) $p H$ = $3$

To Find : Electrode Potential = $E$ = ?

We know that,

$E$ = $E^{\circ}$ - $\frac{0.059}{2}$ $l o g$ [ $H$ $^{+}$ $^{2}$ ]

Substituting the given values we get,

$E$ = $1.30$ - $0.059$ (- $p H$ ) ..............(2 and 2 gets cancelled and $l o g [H^{+}]$ is $- (p H)$

$∴$ $E$ = $1.30 - (- 0.059 \times 3)$

$E$ = $1.48$ $v o l t$

Hence the correct option is 'A'.

Suggest Corrections

Similar questions

For the half cell $E^{o} = 1.30 V$ . At $p H$ $= 3$ , electrode potential is:

p H = 2

p H = 2

E_{Q u i n h y d r o n e}

E_{Q u i h y d r o n e}^{o} = 1.30 V

2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}; E^{\circ} = - 1.23 V

J m o l^{- 1} K^{- 1};

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