Consider the following half cell reaction: The reaction was conducted with the ratio The pH value at which the EMF of the half cell will become zero is _______. (nearest integer value). [Given: Standard half cell reduction potential

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Solution

Q.
Consider the following half cell reaction:

${Cr}_{2} O_{7}^{2 -} (a q) + 6 e^{-} + 14 H^{+} (a q) \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$

The reaction was conducted with the ratio $\frac{{[C r^{3 +}]}^{2}}{[C r_{2} O_{7}^{2 -}]} = 10^{- 6} .$ The pH value at which the EMF of the half cell will become zero is _______. (nearest integer value).

[Given: Standard half cell reduction potential $E_{{Cr}_{2} O_{7}^{2 -}, H^{+} / C r^{3 +}}^{\circ} = 1.33 V, \frac{2.303 R T}{F} = 0.059 V$ [2025]

Ans.
(10)

By Nernst equation:

$E_{reaction} = E_{reaction}^{\circ} - \frac{0.059}{n} \log Q$

$0 = 1.33 - \frac{0.059}{6} \log \frac{{[C r^{3 +}]}^{2}}{[C r_{2} O_{7}^{2 -}] {[H^{+}]}^{14}}$

$0 = 1.33 - \frac{0.059}{6} \log \frac{10^{- 6}}{{[H^{+}]}^{14}}$

$\frac{1.33 \times 6}{0.059} = \log \frac{10^{- 6}}{{[H^{+}]}^{14}}$

$135.254 = \log 10^{- 6} - \log {[H^{+}]}^{14}$

$135.254 = - 6 + 14 (- \log [H^{+}])$

$141.254 = 14 pH$

$pH = \frac{141.254}{14} = 10.09$

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