CBSE 2025, Set 1

Question

Calculate the potential for the half-cell containing 0.01 M K$_2$Cr$_2$O$_7$(aq), 0.01 M Cr$^{3+}$(aq) and $1.0 	imes 10^{-4}$ M H$^+$(aq).

The half cell reaction is:
$$	ext{Cr}_2	ext{O}_7^{2-}(aq) + 14	ext{H}^+(aq) + 6e^- ightarrow 2	ext{Cr}^{3+}(aq) + 7	ext{H}_2	ext{O}(l)$$

Standard electrode potential E° = 1.33 V.  [Given: log 10 = 1]

Edvaya · Accepted Answer

The correct answer is Subjective. Given concentrations: $[	ext{Cr}_2	ext{O}_7^{2-}] = 10^{-2}$ M, $[	ext{Cr}^{3+}] = 10^{-2}$ M, $[	ext{H}^+] = 10^{-4}$ M **Nernst equation** ($n = 6$): $$E = E° - \frac{0.059}{6}\log\frac{[	ext{Cr}^{3+}]^2}{[	ext{Cr}_2	ext{O}_7^{2-}][	ext{H}^+]^{14}}$$ $$= 1.33 - \frac{0.059}{6}\log\frac{(10^{-2})^2}{(10^{-2})(10^{-4})^{14}}$$ $$= 1.33 - \frac{0.059}{6}\log\frac{10^{-4}}{10^{-2} 	imes 10^{-56}} = 1.33 - \frac{0.059}{6}\log 10^{54}$$ $$= 1.33 - \frac{0.059 	imes 54}{6} = 1.33 - 0.531 = \boxed{0.799 	ext{ V}}$$

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