Question

The EMF of the cell at $298\ \text{K}$ is $Mg_{(s)}|Mg^{2+}(aq,\ 0.10\ \text{M})||Ag^{+}(aq,\ 0.001\ \text{M})|Ag_{(s)}$. (Given: $E^{0}_{cell}=3.17\ \text{V}$ and $\frac{2.303RT}{F}=0.06\ \text{V}$)

• A. 3.32 V
• B. 2.96 V
• C. 3.02 V
• D. 3.17 V
• E. 3.47 V

Hint:

Don't forget to square the concentration of $Ag^{+}$ because of the $2Ag^{+}$ stoichiometry.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires the use of the Nernst equation to calculate the electromotive force (EMF) of an electrochemical cell under non-standard conditions (i.e., concentrations are not 1 M).
Step 2: Key Formula or Approach:
1. Nernst Equation: \[ E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log_{10}(Q) \] where \( E_{cell} \) is the cell potential under non-standard conditions, \( E^\circ_{cell} \) is the standard cell potential, n is the number of moles of electrons transferred in the balanced reaction, F is Faraday's constant, and Q is the reaction quotient. 2. Cell Reaction: We need to write the balanced overall cell reaction to determine n and Q. - Anode (Oxidation): Mg\(_{(s)}\) \( \rightarrow \) Mg\(^{2+}\)\(_{(aq)}\) + 2e\(^-\) - Cathode (Reduction): Ag\(^+\)\(_{(aq)}\) + e\(^-\) \( \rightarrow \) Ag\(_{(s)}\) To balance the electrons, we multiply the cathode half-reaction by 2: - 2Ag\(^+\)\(_{(aq)}\) + 2e\(^-\) \( \rightarrow \) 2Ag\(_{(s)}\) Overall reaction: Mg\(_{(s)}\) + 2Ag\(^+\)\(_{(aq)}\) \( \rightarrow \) Mg\(^{2+}\)\(_{(aq)}\) + 2Ag\(_{(s)}\) 3. From the balanced reaction, the number of electrons transferred is n=2. 4. The reaction quotient, Q, is: \[ Q = \frac{[\text{Products}]}{[\text{Reactants}]} = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^{+}]^2} \] (Solids Mg and Ag have activity = 1 and are omitted). Step 3: Detailed Explanation:
We are given: - \( E^\circ_{cell} = 3.17 \) V - \( \frac{2.303RT}{F} = 0.06 \) V - [Mg\(^{2+}\)] = 0.10 M = \( 10^{-1} \) M - [Ag\(^+\)] = 0.001 M = \( 10^{-3} \) M - n = 2 First, calculate the reaction quotient, Q: \[ Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^{+}]^2} = \frac{10^{-1}}{(10^{-3})^2} = \frac{10^{-1}}{10^{-6}} = 10^{-1 - (-6)} = 10^5 \] Now, substitute all values into the Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log_{10}(Q) \] \[ E_{cell} = 3.17 - \frac{0.06}{2} \log_{10}(10^5) \] Since \( \log_{10}(10^5) = 5 \): \[ E_{cell} = 3.17 - (0.03) \times 5 \] \[ E_{cell} = 3.17 - 0.15 \] \[ E_{cell} = 3.02 \text{ V} \] Step 4: Final Answer:
The EMF of the cell is 3.02 V.