CBSE Class 12 Chemistry: Electrochemistry Intext Questions Solutions

The standard electrode potential of the Mg$^{2+}$ | Mg system can be determined by constructing an electrochemical cell using the standard hydrogen electrode (SHE) as a reference. The SHE consists of platinum electrode in contact with H$^{+}$ ions (1 M concentration) and hydrogen gas at 1 bar pressure, and its standard electrode potential is defined as 0 V.

To measure the standard electrode potential of Magnesium, a cell is set up where the Magnesium electrode (in 1 M Mg$^{2+}$ solution) acts as the anode and the standard hydrogen electrode acts as the cathode. The cell can be represented as:

$$Mg\; |\; Mg^\{2+\}(aq,\; 1M)\; ||\; H^\{+\}(aq,\; 1M)\; |\; H\_\{2\}(g,\; 1\; bar),\; Pt\_\{(s)\}$$

The electromotive force (emf) of this cell is then measured. The emf of the cell is calculated using the formula:

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

In this setup, the SHE is the cathode, so $E^\{\backslash Theta\}\_\{cathode\}\; =\; 0$ V. The Magnesium electrode is the anode, so $E^\{\backslash Theta\}\_\{anode\}$ is the standard electrode potential of Magnesium that we want to determine.

Therefore, the measured emf of the cell is:

$$E^\{\backslash Theta\}\_\{cell\}\; =\; 0\; -\; E^\{\backslash Theta\}\_\{Mg^\{2+\}/Mg\}$$

This implies that the standard electrode potential of the Magnesium electrode is the negative of the measured cell emf: $E^\{\backslash Theta\}\_\{Mg^\{2+\}/Mg\}\; =\; -E^\{\backslash Theta\}\_\{cell\}$.

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 solution. Zinc will get oxidized to zinc ions, and copper ions will get reduced to copper metal, which will deposit on the zinc pot.

The chemical reaction that occurs is:

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

This reaction will consume the copper sulphate solution and corrode the zinc pot, making it unsuitable for storage.

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:

$$Fe^\{2+\}\; \backslash longrightarrow\; Fe^\{3+\}\; +\; e^\{-\}$$

The standard electrode potential for this half-reaction is $E^\{\backslash Theta\}\_\{Fe^\{3+\}/Fe^\{2+\}\}\; =\; +0.77$ V.

For a substance to oxidize $Fe^\{2+\}$ to $Fe^\{3+\}$, it must be a stronger oxidizing agent than $Fe^\{3+\}$. This means the substance must have a standard reduction potential greater than +0.77 V. When such a substance acts as the oxidizing agent (gets reduced), it will cause the oxidation of $Fe^\{2+\}$.

Consulting a standard table of electrode potentials, three substances that have standard reduction potentials greater than +0.77 V and can therefore oxidize ferrous ions are:

1. Fluorine (F$_{2}$): $E^\{\backslash Theta\}\_\{F\_\{2\}/F^\{-\}\}\; =\; +2.87$ V
2. Chlorine (Cl$_{2}$): $E^\{\backslash Theta\}\_\{Cl\_\{2\}/Cl^\{-\}\}\; =\; +1.36$ V
3. Oxygen (O$_{2}$) in acidic medium: $E^\{\backslash Theta\}\_\{O\_\{2\}/H\_\{2\}O\}\; =\; +1.23$ V

These substances can readily accept electrons from $Fe^\{2+\}$, causing its oxidation to $Fe^\{3+\}$.

The potential of a hydrogen electrode is determined by the concentration of hydrogen ions ($H^\{+\}$) in the solution. The half-reaction for a hydrogen electrode is:

$$H^\{+\}\; (aq)\; +\; e^\{-\}\; \backslash longrightarrow\; \backslash frac\{1\}\{2\}\; H\_\{2\}\; (g)$$

We are given that the pH of the solution is 10.

The relationship between pH and hydrogen ion concentration is $pH\; =\; -\backslash log[H^\{+\}]$. Therefore, the hydrogen ion concentration is:

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

The potential of the electrode can be calculated using the Nernst equation. For the hydrogen electrode, the number of electrons transferred ($n$) is 1.

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

At 25°C (298 K), the equation can be simplified using $\backslash frac\{RT\}\{F\}\; \backslash approx\; 0.0591$ V and $\backslash ln\; x\; =\; 2.303\; \backslash log\; x$:

$$E\_\{H^\{+\}/H\_\{2\}\}\; =\; E^\{\backslash Theta\}\_\{H^\{+\}/H\_\{2\}\}\; -\; \backslash frac\{0.0591\}\{1\}\backslash log\backslash frac\{1\}\{[H^\{+\}]\}$$

The standard electrode potential for the hydrogen electrode ($E^\{\backslash Theta\}\_\{H^\{+\}/H\_\{2\}\}$) is 0 V by definition.

Substituting the values:

$$E\_\{H^\{+\}/H\_\{2\}\}\; =\; 0\; -\; \backslash frac\{0.0591\}\{1\}\backslash log\backslash frac\{1\}\{10^\{-10\}\}$$

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

Using the logarithm property $\backslash log(a^b)\; =\; b\; \backslash log\; a$:

$$E\_\{H^\{+\}/H\_\{2\}\}\; =\; -0.0591\; \backslash times\; 10$$

$$E\_\{H^\{+\}/H\_\{2\}\}\; =\; -0.591\; \backslash text\{\; V\}$$

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

This question appears to be incomplete as the specific reaction for which the emf needs to be calculated is not provided in the source text. To calculate the emf of a cell, the overall cell reaction, the standard electrode potentials of the involved half-cells, and potentially the concentrations of the reacting species (if not standard conditions) are required.

Generally, the emf of a cell is calculated using the Nernst equation for non-standard conditions or simply by subtracting the anode potential from the cathode potential under standard conditions ($E^\{\backslash Theta\}\_\{cell\}\; =\; E^\{\backslash Theta\}\_\{cathode\}\; -\; E^\{\backslash Theta\}\_\{anode\}$).

If the reaction were provided, one would identify the oxidation and reduction half-reactions, find their standard electrode potentials from a table, and then calculate the standard cell potential. For non-standard conditions, the Nernst equation would be applied.