NCERT · Class 12 · Chemistry · ElectrochemistryCalculate the EMF of the cell in which the following reaction takes place: $Ni(s) + 2Ag^+(0.002 M) \rightarrow Ni^{2+}(0.160 M) + 2Ag(s)$\nGiven that $E^0{cell} = 1.05 V$. (Use Nernst equation, $F = 96500 C mol^{-1}$, $log 2 = 0.3010$, $log 10 = 1$)
Step 1: Write the Nernst Equation for the Cell Reaction\nThe given cell reaction is:
$Ni(s) + 2Ag^+(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)$ \nThe number of electrons transferred in the reaction, $n = 2$. \nAccording to the Nernst equation at $298 K$: $$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Ag^+]^2}$$
Step 2: Identify Given Values
- $E^0_{cell} = 1.05 \text{ V}$
- $[Ni^{2+}] = 0.160 \text{ M} = 1.6 \times 10^{-1} \text{ M}$
- $[Ag^+] = 0.002 \text{ M} = 2 \times 10^{-3} \text{ M}$
- $n = 2$
Step 3: Substitute Values into the Nernst Equation
$$E_{cell} = 1.05 - \frac{0.0591}{2} \log \frac{0.160}{(0.002)^2}$| \nLet us simplify the logarithmic term: $$\frac{0.160}{(0.002)^2} = \frac{0.160}{0.000004} = \frac{160 \times 10^{-3}}{4 \times 10^{-6}} = 40 \times 10^3 = 4 \times 10^4$$ \nNow substitute back into the equation: $$E_{cell} = 1.05 - \frac{0.0591}{2} \log (4 \times 10^4)$$ $$E_{cell} = 1.05 - 0.02955 \times (\log 4 + \log 10^4)$$\nSince $\log 4 = \log (2^2) = 2 \log 2 = 2 \times 0.3010 = 0.6020$ and $\log 10^4 = 4$: $$\log (4 \times 10^4) = 0.6020 + 4 = 4.6020$$
Step 4: Final Calculation
$$E_{cell} = 1.05 - 0.02955 \times 4.6020$$ $$E_{cell} = 1.05 - 0.1360$$ $$E_{cell} = 0.914 \text{ V}$$
Answer: The EMF of the cell is $0.914 \text{ V}$.