Question

Find the emf of the reaction at 298K
\( \text{Ni}_{(s)} + 2\text{Ag}^+_{2\times 10^{-3}\text{M}} \rightarrow \text{Ni}^{+2}_{4\times 10^{-2}\text{M}} + 2\text{Ag}_{(s)} \)
\( (\text{E}^0_{\text{cell}} = 1.5 \text{ at } 298\text{K}) \)

Hint:

Always remember to square the concentration of the silver ion \( [\text{Ag}^+] \) in the reaction quotient \( Q \), as its stoichiometric coefficient in the balanced equation is 2. Missing this power is the most common error in Nernst equation problems.

Answer 1

Step 1: Understanding the Concept:
The electromotive force (EMF) of a cell under non-standard conditions is determined using the Nernst equation, which correlates the cell potential with the standard potential and the reaction quotient.
Step 2: Key Formula or Approach:
The Nernst equation at 298 K is:
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log_{10} Q \] Where \( n \) is the number of moles of electrons transferred, and \( Q \) is the reaction quotient.
Step 3: Detailed Explanation:
From the given cell reaction:
\[ \text{Ni}_{(s)} + 2\text{Ag}^+_{(aq)} \rightarrow \text{Ni}^{+2}_{(aq)} + 2\text{Ag}_{(s)} \] The number of electrons transferred \( n = 2 \).
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Ni}^{+2}]}{[\text{Ag}^+]^2} \] Substitute the given concentrations into the expression for \( Q \):
\[ Q = \frac{4 \times 10^{-2}}{(2 \times 10^{-3})^2} \] Calculate the denominator:
\[ (2 \times 10^{-3})^2 = 4 \times 10^{-6} \] Now, calculate \( Q \):
\[ Q = \frac{4 \times 10^{-2}}{4 \times 10^{-6}} = 10^4 \] Now, substitute \( E^0_{\text{cell}} = 1.5 \text{ V} \), \( n = 2 \), and \( Q = 10^4 \) into the Nernst equation:
\[ E_{\text{cell}} = 1.5 - \frac{0.0591}{2} \log_{10}(10^4) \] Since \( \log_{10}(10^4) = 4 \):
\[ E_{\text{cell}} = 1.5 - \frac{0.0591}{2} \times 4 \] \[ E_{\text{cell}} = 1.5 - (0.0591 \times 2) \] \[ E_{\text{cell}} = 1.5 - 0.1182 = 1.3818 \text{ V} \] Rounding to two decimal places, we get 1.38 V.
Step 4: Final Answer:
The emf of the cell is 1.38 V.