CBSE Class 12 Chemistry Chapter 3: 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 or its reactivity. A more reactive metal can displace a less reactive metal from its salt solution. The general order of reactivity for the given metals, from most reactive to least reactive, is Mg > Al > Zn > Fe > Cu. Therefore, they will displace each other in this order. For instance, Mg can displace Al, Zn, Fe, and Cu from their respective salt solutions, while Cu cannot displace any of the other metals.

The order of displacement is: Mg, Al, Zn, Fe, Cu.

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

The given standard electrode potentials are:

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

Arranging these potentials in increasing order:

Since reducing power increases with decreasing electrode potential, the order of increasing reducing power is the reverse of the order of increasing electrode potential.

Therefore, the increasing order of reducing power is: Ag < Hg < Cr < Mg < K.

Solution: The given reaction is a redox reaction where zinc is oxidized to zinc ions and silver ions are reduced to silver metal. This reaction can occur in a galvanic cell.

The overall reaction is: $Zn(s)\; +\; 2Ag^\{+\}(aq)\; \backslash rightarrow\; Zn^\{2+\}(aq)\; +\; 2Ag(s)$

The standard cell notation for this galvanic cell is:

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

1. Which of the electrode is negatively charged? In a galvanic cell, the anode is the electrode where oxidation occurs, and it is the source of electrons. Therefore, the anode is negatively charged. In this reaction, zinc is oxidized ($Zn\; \backslash rightarrow\; Zn^\{2+\}\; +\; 2e^\{-\}$), so the Zn electrode is negatively charged.
2. The carriers of the current in the cell. Within the electrolyte solutions of the cell, the current is carried by the movement of ions (cations and anions). In the external circuit, the current is carried by the flow of electrons from the anode (negative terminal) to the cathode (positive terminal). So, electrons flow from zinc to silver in the external circuit.
3. Individual reaction at each electrode. At the anode (oxidation): $$Zn\_\{(s)\}\; \backslash longrightarrow\; Zn^\{2+\}\_\{(aq)\}\; +\; 2e^\{-\}$$ At the cathode (reduction): $$Ag^\{+\}\_\{(aq)\}\; +\; e^\{-\}\; \backslash longrightarrow\; Ag\_\{(s)\}$$ To balance the electrons, the reduction half-reaction must be multiplied by 2: $2Ag^\{+\}\_\{(aq)\}\; +\; 2e^\{-\}\; \backslash longrightarrow\; 2Ag\_\{(s)\}$. The overall reaction is obtained by summing the balanced half-reactions.

Solution: To calculate the standard cell potential ($E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}$), standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash theta\}$), and equilibrium constant ($K\_\{\backslash text\{eq\}\}$), we need the standard reduction potentials for the involved half-cells. Standard reduction potentials are typically provided or can be looked up.

We will use the following standard reduction potentials:

• $E^\{\backslash ominus\}\_\{Cr^\{3+\}/Cr\}\; =\; -0.74\; \backslash text\{\; V\}$
• $E^\{\backslash ominus\}\_\{Cd^\{2+\}/Cd\}\; =\; -0.40\; \backslash text\{\; V\}$
• $E^\{\backslash ominus\}\_\{Fe^\{3+\}/Fe^\{2+\}\}\; =\; +0.77\; \backslash text\{\; V\}$
• $E^\{\backslash ominus\}\_\{Ag^\{+\}/Ag\}\; =\; +0.80\; \backslash text\{\; V\}$

Part 1: $2Cr(s)\; +\; 3Cd^\{2+\}(aq)\; \backslash rightarrow\; 2Cr^\{3+\}(aq)\; +\; 3Cd$ In this reaction, Chromium (Cr) is oxidized ($Cr\; \backslash rightarrow\; Cr^\{3+\}$) and Cadmium ions ($Cd^\{2+\}$) are reduced ($Cd^\{2+\}\; \backslash rightarrow\; Cd$). Thus, Cr acts as the anode and Cd acts as the cathode.

• Anode (Oxidation): $Cr\; \backslash rightarrow\; Cr^\{3+\}\; +\; 3e^\{-\}$ (Standard reduction potential $E^\{\backslash ominus\}\_\{Cr^\{3+\}/Cr\}\; =\; -0.74\; \backslash text\{\; V\}$)
• Cathode (Reduction): $Cd^\{2+\}\; +\; 2e^\{-\}\; \backslash rightarrow\; Cd$ (Standard reduction potential $E^\{\backslash ominus\}\_\{Cd^\{2+\}/Cd\}\; =\; -0.40\; \backslash text\{\; V\}$)

The number of electrons transferred, n, is 6 (LCM of 3 and 2). The standard cell potential is calculated as:

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

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

Now, we calculate the standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash theta\}$) using the formula $\backslash Delta\_r\; G^\{\backslash theta\}\; =\; -nFE^\{\backslash Theta\}\_\{\backslash text\{cell\}\}$, where $F\; =\; 96485\; \backslash text\{\; C/mol\}$.

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

$\backslash Delta\_r\; G^\{\backslash theta\}\; =\; -196809.3\; \backslash text\{\; J/mol\}\; \backslash approx\; -196.81\; \backslash text\{\; kJ/mol\}$

The equilibrium constant ($K\_\{\backslash text\{eq\}\}$) is calculated using the formula $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{RT\}\{nF\}\; \backslash ln\; K\_\{\backslash text\{eq\}\}$ or at 298 K, $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{0.0591\}\{n\}\; \backslash log\; K\_\{\backslash text\{eq\}\}$.

$0.34\; \backslash text\{\; V\}\; =\; \backslash frac\{0.0591\; \backslash text\{\; V\}\}\{6\}\; \backslash log\; K\_\{\backslash text\{eq\}\}$

$\backslash log\; K\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{0.34\; \backslash times\; 6\}\{0.0591\}\; \backslash approx\; \backslash frac\{2.04\}\{0.0591\}\; \backslash approx\; 34.517$

$K\_\{\backslash text\{eq\}\}\; =\; 10^\{34.517\}\; \backslash approx\; 3.3\; \backslash times\; 10^\{34\}$

Part 2: $Fe^\{2+\}(aq)\; +\; Ag^\{+\}(aq)\; \backslash rightarrow\; Fe^\{3+\}(aq)\; +\; Ag(s)$ In this reaction, $Fe^\{2+\}$ is oxidized to $Fe^\{3+\}$, and $Ag^\{+\}$ is reduced to Ag. Thus, Fe acts as the anode and Ag acts as the cathode.

• Anode (Oxidation): $Fe^\{2+\}\; \backslash rightarrow\; Fe^\{3+\}\; +\; e^\{-\}$ (Standard reduction potential $E^\{\backslash ominus\}\_\{Fe^\{3+\}/Fe^\{2+\}\}\; =\; +0.77\; \backslash text\{\; V\}$)
• Cathode (Reduction): $Ag^\{+\}\; +\; e^\{-\}\; \backslash rightarrow\; Ag$ (Standard reduction potential $E^\{\backslash ominus\}\_\{Ag^\{+\}/Ag\}\; =\; +0.80\; \backslash text\{\; V\}$)

The number of electrons transferred, n, is 1. The standard cell potential is calculated as:

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

$E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; (+0.80\; \backslash text\{\; V\})\; -\; (+0.77\; \backslash text\{\; V\})\; =\; +0.03\; \backslash text\{\; V\}$

Now, we calculate the standard Gibbs energy change ($\backslash Delta\_r\; G^\{\backslash theta\}$).

$\backslash Delta\_r\; G^\{\backslash theta\}\; =\; -nFE^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; -(1\; \backslash text\{\; mol\})\; \backslash times\; (96485\; \backslash text\{\; C/mol\})\; \backslash times\; (0.03\; \backslash text\{\; V\})$

$\backslash Delta\_r\; G^\{\backslash theta\}\; =\; -2894.55\; \backslash text\{\; J/mol\}\; \backslash approx\; -2.89\; \backslash text\{\; kJ/mol\}$

The equilibrium constant ($K\_\{\backslash text\{eq\}\}$) is calculated using $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; \backslash frac\{0.0591\}\{n\}\; \backslash log\; K\_\{\backslash text\{eq\}\}$.

$0.03\; \backslash text\{\; V\}\; =\; \backslash frac\{0.0591\; \backslash text\{\; V\}\}\{1\}\; \backslash log\; K\_\{\backslash text\{eq\}\}$

$\backslash log\; K\_\{\backslash text\{eq\}\}\; =\; \backslash frac\{0.03\}\{0.0591\}\; \backslash approx\; 0.5076$

$K\_\{\backslash text\{eq\}\}\; =\; 10^\{0.5076\}\; \backslash approx\; 3.22$

Summary of Results:

1. For the reaction $2Cr(s)\; +\; 3Cd^\{2+\}(aq)\; \backslash rightarrow\; 2Cr^\{3+\}(aq)\; +\; 3Cd$: $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; +0.34\; \backslash text\{\; V\}$, $\backslash Delta\_r\; G^\{\backslash theta\}\; \backslash approx\; -196.81\; \backslash text\{\; kJ/mol\}$, $K\_\{\backslash text\{eq\}\}\; \backslash approx\; 3.3\; \backslash times\; 10^\{34\}$.
2. For the reaction $Fe^\{2+\}(aq)\; +\; Ag^\{+\}(aq)\; \backslash rightarrow\; Fe^\{3+\}(aq)\; +\; Ag(s)$: $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; +0.03\; \backslash text\{\; V\}$, $\backslash Delta\_r\; G^\{\backslash theta\}\; \backslash approx\; -2.89\; \backslash text\{\; kJ/mol\}$, $K\_\{\backslash text\{eq\}\}\; \backslash approx\; 3.22$.