At pH -2- E Quinhydrone -1-30 V-E Quinhydrone will be-A- 1-30 VB- 1-36 VC- 1-42 VD- 1-20 V

The correct option is C 1.42 VE_quinohydrone=1.3 0.0591/2log[H^+]^2 =1.3 0.0591/2× 4 =1.42V ...

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Properties of Water

At pH =2, E Q...

Question

At $p H$ = 2, $E_{Q u i n h y d r o n e}^{\circ} = 1.30 V$ ,
$E_{Q u i n h y d r o n e}$ will be:

1.36 V

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1.30 V

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1.42 V

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1.20 V

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Solution

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p H = 2, E_{Quinhydrone}

E_{Quinhydrone}^{o} = 1.30 V

p H = 2

p H = 2, E_{Q u i n h y d r o n e}

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

1 M

Q H_{2} \to Q + 2 H^{+} + 2 e^{-}

Q H_{2}

Q

\frac{2.303 R T}{F} = 0.06

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

The single electrode potential of chloride electrode of [Cl^‑] = 0.1 M is (E^oCl₂/Cl^‑ is 1.36 V)

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