CBSE Class 12 Chemistry Chapter 5: Electrochemistry NCERT Solutions

The standard electrode potential of the Mg²⁺ | Mg system can be determined by constructing an electrochemical cell where the magnesium electrode acts as the anode and the standard hydrogen electrode (SHE) serves as the cathode. The SHE is represented as Pt_(s), H_2(g) (1 atm) | H⁺_(aq) (1 M).

The cell can be represented as: Mg | Mg²⁺(aq, 1M) || H⁺(aq, 1M) | H₂(g, 1 bar), Pt_(s)

The electromotive force (emf) of this cell is then measured. According to the formula for cell potential, $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash Theta\}\_\{\backslash text\{cathode\}\}\; -\; E^\{\backslash Theta\}\_\{\backslash text\{anode\}\}$. In this setup, the SHE is the cathode, and its standard electrode potential ($E^\{\backslash Theta\}\_\{\backslash text\{cathode\}\}$) is defined as 0 V. The magnesium electrode is the anode ($E^\{\backslash Theta\}\_\{\backslash text\{anode\}\}$).

Therefore, the measured emf of the cell is equal to the standard electrode potential of the magnesium electrode: $E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; 0\; -\; E^\{\backslash Theta\}\_\{\backslash text\{Mg^\{2+\}/Mg\}\}$, which means $E^\{\backslash Theta\}\_\{\backslash text\{Mg^\{2+\}/Mg\}\}\; =\; -E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}$.

No, copper sulphate solutions cannot be stored in a zinc pot. This is because zinc is a more reactive metal than copper. According to the reactivity series, a more reactive metal can displace a less reactive metal from its salt solution.

When copper sulphate solution is placed in a zinc pot, the zinc metal will react with the copper sulphate. Zinc will displace copper ions from the solution, forming zinc sulphate and solid copper. The chemical reaction is:

$$Zn\_\{(s)\}\; +\; CuSO\_\{4(aq)\}\; \backslash longrightarrow\; ZnSO\_\{4(aq)\}\; +\; Cu\_\{(s)\}$$

This reaction will consume the copper sulphate solution and coat the zinc pot with copper, rendering the storage impossible and contaminating the solution.

Ferrous ions ($Fe^\{2+\}$) can be oxidized to ferric ions ($Fe^\{3+\}$) if a substance with a sufficiently high reduction potential is present. The oxidation half-reaction is: $Fe^\{2+\}\; \backslash longrightarrow\; Fe^\{3+\}\; +\; e^\{-1\}$, with a standard electrode potential ($E^\{\backslash Theta\}$) of -0.77 V.

For a substance to oxidize $Fe^\{2+\}$, it must act as an oxidizing agent, meaning it gets reduced. This requires the substance to have a standard reduction potential greater than the standard oxidation potential of the $Fe^\{2+\}/Fe^\{3+\}$ couple (which is the negative of the reduction potential, i.e., +0.77 V). In other words, the substance must be a stronger oxidizing agent than $Fe^\{3+\}$.

Consulting a standard table of electrode potentials, substances with reduction potentials greater than +0.77 V can oxidize ferrous ions. Three such substances are:

1. Fluorine ($F\_2$): $E^\{\backslash Theta\}\; =\; +2.87\; \backslash text\{\; V\}$
2. Chlorine ($Cl\_2$): $E^\{\backslash Theta\}\; =\; +1.36\; \backslash text\{\; V\}$
3. Oxygen ($O\_2$) in acidic solution: $E^\{\backslash Theta\}\; =\; +1.23\; \backslash text\{\; V\}$

The half-reaction for a hydrogen electrode is: $$ H^\{+\}\_\{(aq)\}\; +\; e^\{-\}\; \backslash longrightarrow\; \backslash frac\{1\}\{2\}\; H\_\{2(g)\}$$

Given that the pH of the solution is 10. The pH is defined as $pH\; =\; -\backslash log[H^\{+\}]$. Therefore, the concentration of hydrogen ions is:

$[H^\{+\}]\; =\; 10^\{-pH\}\; =\; 10^\{-10\}\; \backslash text\{\; M\}$

The potential of the hydrogen electrode under non-standard conditions can be calculated using the Nernst equation:

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}^\{\backslash Theta\}\; -\; \backslash frac\{RT\}\{nF\}\backslash ln\backslash frac\{1\}\{[H^\{+\}]\}$$

At 25°C (298 K), the term $\backslash frac\{RT\}\{nF\}\backslash ln$ can be simplified to $\backslash frac\{0.0591\}\{n\}\backslash log$. For the hydrogen electrode, $n=1$ (number of electrons transferred). The standard electrode potential for the hydrogen electrode ($E^\{\backslash Theta\}\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}$) is 0 V by convention.

Substituting the values into the Nernst equation:

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; 0\; -\; \backslash frac\{0.0591\}\{1\}\backslash log\backslash frac\{1\}\{[H^\{+\}]\}$$

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; -0.0591\; \backslash log\backslash frac\{1\}\{10^\{-10\}\}$$

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; -0.0591\; \backslash log(10^\{10\})$$

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; -0.0591\; \backslash times\; 10$$

$$E\_\{\backslash left(H^\{+\}/\backslash frac\{1\}\{2\}H\_\{2\}\backslash right)\}\; =\; -0.591\; \backslash text\{\; V\}$$

Thus, the potential of the hydrogen electrode in a solution with pH 10 is -0.591 V.

To calculate the emf of the cell, we need the standard electrode potentials for the reduction half-reactions involved. The overall reaction needs to be broken down into its oxidation and reduction half-reactions. Let's assume the reaction involves two specific half-cells (the actual reaction details are missing from the source, so a general approach is provided).

For example, if the reaction was:

$$Zn\_\{(s)\}\; +\; Cu^\{2+\}\_\{(aq)\}\; \backslash longrightarrow\; Zn^\{2+\}\_\{(aq)\}\; +\; Cu\_\{(s)\}$$

The oxidation half-reaction is: $Zn\_\{(s)\}\; \backslash longrightarrow\; Zn^\{2+\}\_\{(aq)\}\; +\; 2e^\{-\}$

The reduction half-reaction is: $Cu^\{2+\}\_\{(aq)\}\; +\; 2e^\{-\}\; \backslash longrightarrow\; Cu\_\{(s)\}$

We would then find the standard electrode potentials ($E^\{\backslash Theta\}$) for these half-reactions from a table:

$E^\{\backslash Theta\}\_\{Zn^\{2+\}/Zn\}\; =\; -0.76\; \backslash text\{\; V\}$

$E^\{\backslash Theta\}\_\{Cu^\{2+\}/Cu\}\; =\; +0.34\; \backslash text\{\; V\}$

The standard emf of the cell ($E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}$) is calculated as:

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

In this example, copper is reduced (cathode) and zinc is oxidized (anode):

$$E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; E^\{\backslash Theta\}\_\{Cu^\{2+\}/Cu\}\; -\; E^\{\backslash Theta\}\_\{Zn^\{2+\}/Zn\}$$

$$E^\{\backslash Theta\}\_\{\backslash text\{cell\}\}\; =\; (+0.34\; \backslash text\{\; V\})\; -\; (-0.76\; \backslash text\{\; V\})\; =\; +1.10\; \backslash text\{\; V\}$$

If the reaction involves non-standard conditions (different concentrations or pressures), the Nernst equation would be used to calculate the cell potential ($E\_\{\backslash text\{cell\}\}$).