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QJEEM23Ch03e0: JEE Main 2023 Chemistry - Class 12 JEEM23Ch03e0 Solution: At 298 K, a 1 litre solution containing 10 mmol of and 100 mmol …

At 298 K, a 1 litre solution containing 10 mmol of Cr2O72−\mathrm{C{r_2}O_7^{2 - }} and 100 mmol of Cr3+\mathrm{Cr^{3+}} shows a pH of 3.0.

Given : Cr2O72−→Cr3+ ; E∘=1.330\mathrm{C{r_2}O_7^{2 - } \to C{r^{3 + }}\,;\,E^\circ = 1.330} V

and 2.303 RTF=0.059\mathrm{{{2.303\,RT} \over F} = 0.059} V

The potential for the half cell reaction is x×10−3x\times10^{-3} V. The value of xx is __________

Step-by-Step Verified Solution

amp;Cr2O72−+14H++6e−→2Cr3++7H2Oamp;E=1.33−0.0596log⁡(0.1)2(10−2)(10−3)14amp;E=1.33−0.0596×42=0.917amp;E=917×10−3amp;x=917\begin{aligned}& \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\\\& \mathrm{E}=1.33-\frac{0.059}{6} \log \frac{(0.1)^2}{\left(10^{-2}\right)\left(10^{-3}\right)^{14}} \\\\& \mathrm{E}=1.33-\frac{0.059}{6} \times 42=0.917 \\\\& \mathrm{E}=917 \times 10^{-3} \\\\& \mathrm{x}=917\end{aligned}

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