At pH - 2- EQuinhydrone will be EQuinhydrone - 1-30 V -Assume that the concentration of hydroquinone and quinone is 1 M-QH2 - Q - 2H- - 2e- where QH2 and Q stand for hydroquinone and quinone respectively- Given- 2-303 RTF - 0-06

QH_2 → Q + 2H^+ + 2e^ E = E^o 2.303RTnF log[H^+]^2[Q][QH_2] Give, pH = 2 ⇒ log[H^+] = 2 ⇒ [H^+] = 10^ 2 ⇒ E =1.30 0.062 log(10^ 2)^2(1)(1) ⇒ E = 1.30 0. ...

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Byju's Answer

Standard XII

Chemistry

Solubility Product

At pH = 2, EQ...

Question

At $p H = 2, E_{Q u i n h y d r o n e}$ will be ( $E_{Q u i n h y d r o n e} = 1.30 V)$ [Assume that the concentration of hydroquinone and quinone is $1 M$ ]
$Q H_{2} \to Q + 2 H^{+} + 2 e^{-}$ , where $Q H_{2}$ and $Q$ stand for hydroquinone and quinone respectively. (Given, $\frac{2.303 R T}{F} = 0.06$ )

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Solution

$Q H_{2} \to Q + 2 H^{+} + 2 e^{-}$
$E = E^{o} - \frac{2.303 R T}{n F} l o g \frac{[H^{+}]^{2} [Q]}{[Q H_{2}]}$
Give, $p H = 2 \Rightarrow - l o g [H^{+}] = 2$
$\Rightarrow [H^{+}] = 10^{- 2}$
$\Rightarrow E = 1.30 - \frac{0.06}{2} l o g \frac{(10^{- 2})^{2} (1)}{(1)}$
$\Rightarrow E = 1.30 - \frac{0.06}{2} l o g 10^{- 4}$
$\Rightarrow E = 1.30 - \frac{0.06}{2} (- 4)$
$\Rightarrow E = 1.30 + 0.06 \times 2$
$\Rightarrow E = 1.30 + 0.12$
$\Rightarrow E = 1.42 V$

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At pH=2, EQuinhydrone will be (EQuinhydrone=1.30 V) [Assume that the concentration of hydroquinone and quinone is 1 M] QH2→Q+2H++2e−, where QH2 and Q stand for hydroquinone and quinone respectively. (Given, 2.303RTF=0.06)

QH2→Q+2H++2e− E=Eo−2.303RTnF log[H+]2[Q][QH2] Give, pH=2⇒−log[H+]=2 ⇒[H+]=10−2 ⇒E=1.30−0.062 log(10−2)2(1)(1) ⇒E=1.30−0.062 log10−4 ⇒E=1.30−0.062(−4) ⇒E=1.30+0.06×2 ⇒E=1.30+0.12 ⇒E=1.42 V