JEE MAIN - Chemistry (2024 - 9th April Morning Shift - No. 21)

The standard reduction potentials at $$298 \mathrm{~K}$$ for the following half cells are given below :

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \quad \mathrm{E}^{\circ}=1.33 \mathrm{~V}$$

$$\begin{array}{ll} \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} & \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} & \mathrm{E}^{\circ}=1.40 \mathrm{~V} \end{array}$$

Consider the given electrochemical reactions,

The number of metal(s) which will be oxidized be $$\mathrm{Cr}_2 \mathrm{O}_7^{2-}$$, in aqueous solution is _________.

Answer
3

Explanation

To determine the number of metals that will be oxidized by $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$ in aqueous solution, we need to compare their standard reduction potentials with that of the given reduction potential of $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$.

The reduction reaction for $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$ is:

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6\mathrm{e}^- \rightarrow 2\mathrm{Cr}^{3+}+7\mathrm{H}_2\mathrm{O}, \quad \mathrm{E}^{\circ}=1.33 \mathrm{~V}$$

This means that $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$ has a strong tendency to get reduced (due to its high reduction potential of 1.33 V). Thus, any metal with a lower standard reduction potential than 1.33 V has the potential to be oxidized by $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$.

Given the standard reduction potentials:

$$\begin{array}{ll} \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} & \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} & \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} & \mathrm{E}^{\circ}=1.40 \mathrm{~V} \end{array}$$

Now we compare these reduction potentials with that of $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$:

  • For $$\mathrm{Fe}^{3+}/\mathrm{Fe}$$, $$\mathrm{E}^{\circ}=-0.04 \mathrm{~V}$$ (lower than 1.33 V)
  • For $$\mathrm{Ni}^{2+}/\mathrm{Ni}$$, $$\mathrm{E}^{\circ}=-0.25 \mathrm{~V}$$ (lower than 1.33 V)
  • For $$\mathrm{Ag}^{+}/\mathrm{Ag}$$, $$\mathrm{E}^{\circ}=0.80 \mathrm{~V}$$ (lower than 1.33 V)
  • For $$\mathrm{Au}^{3+}/\mathrm{Au}$$, $$\mathrm{E}^{\circ}=1.40 \mathrm{~V}$$ (higher than 1.33 V)

From the comparison, we see that $$\mathrm{Fe}$$, $$\mathrm{Ni}$$, and $$\mathrm{Ag}$$ have standard reduction potentials lower than 1.33 V, meaning these metals can be oxidized by $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$, but $$\mathrm{Au}$$ cannot.

Therefore, the number of metals that will be oxidized by $$\mathrm{Cr}_2\mathrm{O}_7^{2-}$$ in aqueous solution is: 3.

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