CBSE Class 12 Chemistry Chapter 6: Electrochemistry NCERT Solutions

Solution:

The ability of a metal to displace another metal from its salt solution depends on its position in the electrochemical series. A more reactive metal (higher in the series) can displace a less reactive metal (lower in the series) from its salt solution. The standard electrode potentials indicate the relative reactivity. Metals with more negative standard electrode potentials are more reactive.

The standard electrode potentials for the given metals are approximately:

• Mg: -2.37 V
• Al: -1.66 V
• Zn: -0.76 V
• Fe: -0.44 V
• Cu: +0.34 V

Arranging these metals in decreasing order of their reactivity (or increasing order of their standard electrode potentials) gives:

Mg > Al > Zn > Fe > Cu

Therefore, the order in which these metals displace each other from the solution of their salts is:

Mg, Al, Zn, Fe, Cu

This means Mg can displace Al, Zn, Fe, and Cu; Al can displace Zn, Fe, and Cu; Zn can displace Fe and Cu; and Fe can displace Cu.

Solution:

The reducing power of a metal is its tendency to lose electrons and get oxidized. This tendency is inversely related to its standard electrode potential. A lower (more negative) standard electrode potential indicates a greater tendency to lose electrons, and thus, higher reducing power.

The given standard electrode potentials are:

• $K^\{+\}/K$: -2.93 V
• $Mg^\{2+\}/Mg$: -2.37 V
• $Cr^\{3+\}/Cr$: -0.74 V
• $Hg^\{2+\}/Hg$: +0.79 V
• $Ag^\{+\}/Ag$: +0.80 V

Arranging these potentials in increasing order:

This corresponds to the order:

Since reducing power increases as the standard electrode potential decreases, the metals in increasing order of their reducing power are:

Ag < Hg < Cr < Mg < K

Solution:

The given reaction is a spontaneous redox reaction that can be used to construct a galvanic cell. The reaction involves the oxidation of Zinc (Zn) to Zinc ions (Zn²⁺) and the reduction of Silver ions (Ag⁺) to Silver (Ag).

The cell representation follows the convention: Anode | Anode electrolyte || Cathode electrolyte | Cathode.

The galvanic cell can be depicted as:

$Zn\_\{(s)\}\; |\; Zn^\{2+\}\_\{(aq)\}\; ||\; Ag^\{+\}\_\{(aq)\}\; |\; Ag\_\{(s)\}$

1. Which of the electrode is negatively charged? In a galvanic cell, the anode is the electrode where oxidation occurs. Oxidation involves the release of electrons. Therefore, the anode serves as the source of electrons and is the negatively charged electrode. In this cell, Zinc is oxidized, so the Zn electrode is negatively charged.
2. The carriers of the current in the cell. The carriers of current within the electrolyte of the cell are ions. Cations move towards the cathode, and anions move towards the anode. In the external circuit, the current is carried by the flow of electrons from the anode (Zn) to the cathode (Ag).
3. Individual reaction at each electrode. At the anode (oxidation): $Zn\_\{(s)\}\; \backslash longrightarrow\; Zn^\{2+\}\_\{(aq)\}\; +\; 2e^\{-\}$ At the cathode (reduction): Since the overall reaction shows 2 moles of Ag⁺ being reduced, we need to balance the electrons. The reduction half-reaction is: $Ag^\{+\}\_\{(aq)\}\; +\; e^\{-\}\; \backslash longrightarrow\; Ag\_\{(s)\}$ To balance the electrons lost in the anode reaction (2e^-), this cathode reaction must be multiplied by 2: $2Ag^\{+\}\_\{(aq)\}\; +\; 2e^\{-\}\; \backslash longrightarrow\; 2Ag\_\{(s)\}$

Solution:

To calculate the standard cell potential ($E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}$), standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash ominus\}$), and equilibrium constant (K), we need the standard electrode potentials for the relevant half-reactions. We will use the provided values and standard tables where necessary.

Part 1: $2Cr(s)\; +\; 3Cd^\{2+\}(aq)\; \backslash rightarrow\; 2Cr^\{3+\}(aq)\; +\; 3Cd$

The half-reactions are:

• Oxidation (Anode): $Cr(s)\; \backslash rightarrow\; Cr^\{3+\}(aq)\; +\; 3e^\{-\}$
• Reduction (Cathode): $Cd^\{2+\}(aq)\; +\; 2e^\{-\}\; \backslash rightarrow\; Cd(s)$

From standard tables (or provided context if available), the standard electrode potentials are approximately:

• $E^\{\backslash ominus\}\_\{Cr^\{3+\}/Cr\}\; =\; -0.74\; \backslash text\{\; V\}$
• $E^\{\backslash ominus\}\_\{Cd^\{2+\}/Cd\}\; =\; -0.40\; \backslash text\{\; V\}$

In this reaction, Cr is oxidized and Cd²⁺ is reduced. Therefore:

• Anode: $Cr/Cr^\{3+\}$
• Cathode: $Cd^\{2+\}/Cd$

The standard cell potential is calculated as:

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash ominus\}\_\{\backslash text\{cathode\}\}\; -\; E^\{\backslash ominus\}\_\{\backslash text\{anode\}\}$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash ominus\}\_\{Cd^\{2+\}/Cd\}\; -\; E^\{\backslash ominus\}\_\{Cr^\{3+\}/Cr\}$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; (-0.40\; \backslash text\{\; V\})\; -\; (-0.74\; \backslash text\{\; V\})$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; -0.40\; \backslash text\{\; V\}\; +\; 0.74\; \backslash text\{\; V\}\; =\; +0.34\; \backslash text\{\; V\}$

Now, we calculate the standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash ominus\}$). The number of electrons transferred (n) in the balanced reaction is 6 (since Cr goes from 0 to +3, and Cd goes from +2 to 0, the LCM of 3 and 2 is 6).

$\backslash Delta\_r\; G^\{\backslash ominus\}\; =\; -nFE^\{\backslash ominus\}\_\{\backslash text\{cell\}\}$

Using $F\; =\; 96485\; \backslash text\{\; C\; mol\}^\{-1\}$:

$\backslash Delta\_r\; G^\{\backslash ominus\}\; =\; -(6\; \backslash text\{\; mol\})\; \backslash times\; (96485\; \backslash text\{\; C\; mol\}^\{-1\})\; \backslash times\; (+0.34\; \backslash text\{\; V\})$

$\backslash Delta\_r\; G^\{\backslash ominus\}\; \backslash approx\; -196809.4\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash approx\; -196.81\; \backslash text\{\; kJ\; mol\}^\{-1\}$

Finally, we calculate the equilibrium constant (K). The relationship is:

$\backslash Delta\_r\; G^\{\backslash ominus\}\; =\; -RT\; \backslash ln\; K$

Or, using $E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{RT\}\{nF\}\; \backslash ln\; K$, which at 298 K simplifies to $E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{0.0592\}\{n\}\; \backslash log\; K$.

$0.34\; \backslash text\{\; V\}\; =\; \backslash frac\{0.0592\; \backslash text\{\; V\}\}\{6\}\; \backslash log\; K$

$\backslash log\; K\; =\; \backslash frac\{0.34\; \backslash times\; 6\}\{0.0592\}\; \backslash approx\; \backslash frac\{2.04\}\{0.0592\}\; \backslash approx\; 34.46$

$K\; =\; 10^\{34.46\}\; \backslash approx\; 2.88\; \backslash times\; 10^\{34\}$

Part 2: $Fe^\{2+\}(aq)\; +\; Ag^\{+\}(aq)\; \backslash rightarrow\; Fe^\{3+\}(aq)\; +\; Ag(s)$

The half-reactions are:

• Oxidation (Anode): $Fe^\{2+\}(aq)\; \backslash rightarrow\; Fe^\{3+\}(aq)\; +\; e^\{-\}$
• Reduction (Cathode): $Ag^\{+\}(aq)\; +\; e^\{-\}\; \backslash rightarrow\; Ag(s)$

From standard tables, the standard electrode potentials are approximately:

• $E^\{\backslash ominus\}\_\{Fe^\{3+\}/Fe^\{2+\}\}\; =\; +0.77\; \backslash text\{\; V\}$
• $E^\{\backslash ominus\}\_\{Ag^\{+\}/Ag\}\; =\; +0.80\; \backslash text\{\; V\}$

In this reaction, Fe²⁺ is oxidized and Ag⁺ is reduced. Therefore:

• Anode: $Fe^\{3+\}/Fe^\{2+\}$
• Cathode: $Ag^\{+\}/Ag$

The standard cell potential is calculated as:

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash ominus\}\_\{\backslash text\{cathode\}\}\; -\; E^\{\backslash ominus\}\_\{\backslash text\{anode\}\}$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash ominus\}\_\{Ag^\{+\}/Ag\}\; -\; E^\{\backslash ominus\}\_\{Fe^\{3+\}/Fe^\{2+\}\}$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; (+0.80\; \backslash text\{\; V\})\; -\; (+0.77\; \backslash text\{\; V\})$

$E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; +0.03\; \backslash text\{\; V\}$

Now, we calculate the standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash ominus\}$). The number of electrons transferred (n) is 1.

$\backslash Delta\_r\; G^\{\backslash ominus\}\; =\; -nFE^\{\backslash ominus\}\_\{\backslash text\{cell\}\}$

$\backslash Delta\_r\; G^\{\backslash ominus\}\; =\; -(1\; \backslash text\{\; mol\})\; \backslash times\; (96485\; \backslash text\{\; C\; mol\}^\{-1\})\; \backslash times\; (+0.03\; \backslash text\{\; V\})$

$\backslash Delta\_r\; G^\{\backslash ominus\}\; \backslash approx\; -2894.55\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash approx\; -2.89\; \backslash text\{\; kJ\; mol\}^\{-1\}$

Finally, we calculate the equilibrium constant (K) using $E^\{\backslash ominus\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{0.0592\}\{n\}\; \backslash log\; K$ at 298 K.

$0.03\; \backslash text\{\; V\}\; =\; \backslash frac\{0.0592\; \backslash text\{\; V\}\}\{1\}\; \backslash log\; K$

$\backslash log\; K\; =\; \backslash frac\{0.03\}\{0.0592\}\; \backslash approx\; 0.5067$

$K\; =\; 10^\{0.5067\}\; \backslash approx\; 3.21$