CBSE · Class 12 · Chemistry · ElectrochemistryCalculate the EMF of the cell in which the following reaction takes place: $$\text{Mg}(s) + 2\text{Ag}^+(0.0001 \text{ M}) \rightarrow \text{Mg}^{2+}(0.130 \text{ M}) + 2\text{Ag}(s)$$\nGiven that $E^0{\text{cell}} = +3.17 \text{ V}$. (Use $\log(1.3 \times 10^7) = 7.114$ or calculate appropriately).

To calculate the EMF of the cell, we use the Nernst equation:

$$E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} \right)$$

Step 1: Identify the given values from the problem and reaction.

• $E^0_{\text{cell}} = +3.17 \text{ V}$
• $[\text{Mg}^{2+}] = 0.130 \text{ M} = 1.3 \times 10^{-1} \text{ M}$
• $[\text{Ag}^+] = 0.0001 \text{ M} = 1.0 \times 10^{-4} \text{ M}$
• $n = 2$ (since 2 electrons are transferred in the overall redox reaction: $\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$ and $2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$)

Step 2: Substitute the values into the Nernst equation. $$E_{\text{cell}} = 3.17 - \frac{0.0591}{2} \log \left( \frac{0.130}{(0.0001)^2} \right)$$

Step 3: Simplify the concentration term (reaction quotient, $Q$). $$Q = \frac{0.130}{(10^{-4})^2} = \frac{1.3 \times 10^{-1}}{10^{-8}} = 1.3 \times 10^7$$

Step 4: Calculate the logarithmic value. $$\log(1.3 \times 10^7) = \log(1.3) + \log(10^7) = 0.1139 + 7 = 7.1139 \approx 7.11$$

Step 5: Complete the arithmetic calculation. $$E_{\text{cell}} = 3.17 - \frac{0.0591}{2} \times 7.11$$ $$E_{\text{cell}} = 3.17 - 0.02955 \times 7.11$$ $$E_{\text{cell}} = 3.17 - 0.2101$$ $$E_{\text{cell}} = 2.9599 \text{ V} \approx 2.96 \text{ V}$$

Answer:\nThe EMF of the given cell is $2.96 \text{ V}$.