The standard reduction potentials at 298 ~K for the following half cells are given below : Cr_2 O_7^2-+14 H^++6 e^- arrow 2 Cr^3++7 H_2 O, \quad E^\circ=1.33 ~V\beginarrayll Fe^3+(aq)+3 e^- arrow Fe & E^\circ=-0.04 ~V Ni^2+(aq)+2 e^- arrow Ni & E^\circ=-0.25 ~V Ag^+(aq)+e^- arrow Ag & E^\circ=0.80 ~V Au^3+(aq)+3 e^- arrow Au & E^\circ=1.40 ~V \endarray Consider the given electrochemical reactions, The number of metal(s) which will be oxidized be Cr_2 O_7^2- , in aqueous solution is \underline\hspace2cm .

Electrochemistry PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q21 with Solution

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Correct answer: 3

Oxidizing ability of $\mathrm{Cr_2O_7^{2-}}$

The given reduction half-reaction is:

\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}, \qquad E^\circ = 1.33\,\text{V}

If dichromate oxidizes a metal $M$ , then:

dichromate will undergo reduction at cathode,
metal will undergo oxidation.

For spontaneity,

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{cathode}) - E^\circ_{\text{red}}(\text{anode}) > 0

Here,

E^\circ_{\text{cell}} = 1.33 - E^\circ(M^{n+}/M)

So oxidation of the metal by dichromate is possible if

1.33 > E^\circ(M^{n+}/M)

Check each metal

We are given these standard reduction potentials:

$\mathrm{Fe^{3+} + 3e^- \rightarrow Fe}, \quad E^\circ = -0.04\,V$
$\mathrm{Ni^{2+} + 2e^- \rightarrow Ni}, \quad E^\circ = -0.25\,V$
$\mathrm{Ag^+ + e^- \rightarrow Ag}, \quad E^\circ = 0.80\,V$
$\mathrm{Au^{3+} + 3e^- \rightarrow Au}, \quad E^\circ = 1.40\,V$

Now calculate $E^\circ_{\text{cell}}$ for oxidation by dichromate.

(i) Iron

E^\circ_{\text{cell}} = 1.33 - (-0.04) = 1.37\,V > 0

So Fe will be oxidized.

(ii) Nickel

E^\circ_{\text{cell}} = 1.33 - (-0.25) = 1.58\,V > 0

So Ni will be oxidized.

(iii) Silver

E^\circ_{\text{cell}} = 1.33 - 0.80 = 0.53\,V > 0

So Ag will be oxidized.

(iv) Gold

E^\circ_{\text{cell}} = 1.33 - 1.40 = -0.07\,V < 0

So Au will not be oxidized.

Count the metals oxidized

The metals oxidized by $\mathrm{Cr_2O_7^{2-}}$ are:

\mathrm{Fe,\ Ni,\ Ag}

Thus, the number of metals is:

\boxed{3}

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View written solutionFree

Correct answer: 3

Oxidizing ability of $\mathrm{Cr_2O_7^{2-}}$

The given reduction half-reaction is:

\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}, \qquad E^\circ = 1.33\,\text{V}

If dichromate oxidizes a metal $M$ , then:

dichromate will undergo reduction at cathode,
metal will undergo oxidation.

For spontaneity,

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{cathode}) - E^\circ_{\text{red}}(\text{anode}) > 0

Here,

E^\circ_{\text{cell}} = 1.33 - E^\circ(M^{n+}/M)

So oxidation of the metal by dichromate is possible if

1.33 > E^\circ(M^{n+}/M)

Check each metal

We are given these standard reduction potentials:

$\mathrm{Fe^{3+} + 3e^- \rightarrow Fe}, \quad E^\circ = -0.04\,V$
$\mathrm{Ni^{2+} + 2e^- \rightarrow Ni}, \quad E^\circ = -0.25\,V$
$\mathrm{Ag^+ + e^- \rightarrow Ag}, \quad E^\circ = 0.80\,V$
$\mathrm{Au^{3+} + 3e^- \rightarrow Au}, \quad E^\circ = 1.40\,V$

Now calculate $E^\circ_{\text{cell}}$ for oxidation by dichromate.

(i) Iron

E^\circ_{\text{cell}} = 1.33 - (-0.04) = 1.37\,V > 0

So Fe will be oxidized.

(ii) Nickel

E^\circ_{\text{cell}} = 1.33 - (-0.25) = 1.58\,V > 0

So Ni will be oxidized.

(iii) Silver

E^\circ_{\text{cell}} = 1.33 - 0.80 = 0.53\,V > 0

So Ag will be oxidized.

(iv) Gold

E^\circ_{\text{cell}} = 1.33 - 1.40 = -0.07\,V < 0

So Au will not be oxidized.

Count the metals oxidized

The metals oxidized by $\mathrm{Cr_2O_7^{2-}}$ are:

\mathrm{Fe,\ Ni,\ Ag}

Thus, the number of metals is:

\boxed{3}

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