JEE MAIN - Chemistry (2023 - 24th January Morning Shift - No. 21)
At 298 K, a 1 litre solution containing 10 mmol of $$\mathrm{C{r_2}O_7^{2 - }}$$ and 100 mmol of $$\mathrm{Cr^{3+}}$$ shows a pH of 3.0.
Given : $$\mathrm{C{r_2}O_7^{2 - } \to C{r^{3 + }}\,;\,E^\circ = 1.330}$$V
and $$\mathrm{{{2.303\,RT} \over F} = 0.059}$$ V
The potential for the half cell reaction is $$x\times10^{-3}$$ V. The value of $$x$$ is __________
Answer
917
Explanation
$$
\begin{aligned}
& \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\\\
& \mathrm{E}=1.33-\frac{0.059}{6} \log \frac{(0.1)^2}{\left(10^{-2}\right)\left(10^{-3}\right)^{14}} \\\\
& \mathrm{E}=1.33-\frac{0.059}{6} \times 42=0.917 \\\\
& \mathrm{E}=917 \times 10^{-3} \\\\
& \mathrm{x}=917
\end{aligned}
$$
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