JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the following half cell reaction
\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq})+6 \mathrm{e}^{-}+14 \mathrm{H}^{+}(\mathrm{aq})\) \(\rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_2 \mathrm{O}(\mathrm{l})\)
The reaction was conducted with the ratio of \(\frac{\left[\mathrm{Cr}^{3+}\right]^2}{\left[\mathrm{Cr}_2 \mathrm{O}_7^{2-}\right]}=10^{-6}\). The pH value at which the EMF of the half cell will become zero is ________. (nearest integer value)
[Given : standard half cell reduction potential
\(\left.\mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-}, \mathrm{H}^{+} / \mathrm{Cr}^{3+}}^{\mathrm{o}}=1.33 \mathrm{~V}, \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.059 \mathrm{~V}\right]\)
- A 10
- B 20
- C 30
- D 40
Answer & Solution
Correct Answer
(A) 10
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_{7 \text { (aq) }}^{-2}+14 \mathrm{H}_{(\mathrm{aq})}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{+3}+7 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \mathrm{E}_{\mathrm{R}}=\mathrm{E}_{\mathrm{R}}^0-\frac{0.059}{6} \log…
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