Question

Silver rod is dipped in aqueous solution of \( \mathrm{AgNO_3} \) of unknown concentration and Zn rod is dipped in \( 1 \, \mathrm{M} \) \( \mathrm{ZnSO_4} \) aqueous solution. Both these containers are connected to form a galvanic cell showing emf of \( 1.6 \, \mathrm{V} \). Calculate the value of \( \log_{10}[\mathrm{Ag}^+] \).
\( E^\circ_{\mathrm{Ag^+/Ag(s)} = 0.8 \, \mathrm{V} \qquad E^\circ_{\mathrm{Zn^{2+}/Zn(s)}} = -0.76 \, \mathrm{V} \)
[Use \( \dfrac{2.303RT}{F} = 0.059 \)]

• A. \( \dfrac{5.9}{4} \)
• B. \( \dfrac{4}{5.9} \)
• C. \( \dfrac{2}{5.9} \)
• D. \( \dfrac{8}{5.9} \)

Hint:

In galvanic cell problems, first identify cathode and anode using standard reduction potentials, then write the correct reaction quotient \( Q \) before applying the Nernst equation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves the Nernst equation for a galvanic cell.
The Nernst equation relates the cell EMF to the standard EMF and the reaction quotient \(Q\).
First, identify the anode and cathode based on standard reduction potentials (SRP).
The electrode with the lower SRP acts as the anode (undergoes oxidation).
Anode (oxidation): Zn rod, \(E^\circ_{red} = -0.76\) V
Cathode (reduction): Ag rod, \(E^\circ_{red} = +0.8\) V
Step 2: Key Formula or Approach:
1. Calculate \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\).
2. Write the overall balanced cell reaction and determine \(n\) (number of electrons transferred).
3. Apply the Nernst Equation: \(E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log Q\).
Step 3: Detailed Explanation:
Step 3.1: Standard Cell EMF
\[ E^\circ_{cell} = E^\circ_{Ag^{+}/Ag} - E^\circ_{Zn^{2+}/Zn} = 0.8 - (-0.76) = 1.56 \text{ V} \]
Step 3.2: Overall Cell Reaction
To balance electrons, multiply the Ag half-reaction by 2:
Cathode: \(2Ag^{+}(aq) + 2e^{-} \rightarrow 2Ag(s)\)
Anode: \(Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}\)
Overall: \(Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)\)
Here, \(n = 2\) (2 electrons transferred).
Reaction quotient: \(Q = \frac{[Zn^{2+}]}{[Ag^{+}]^{2}}\)
Step 3.3: Apply Nernst Equation
Given: \(E_{cell} = 1.6\) V, \([Zn^{2+}] = 1\) M.
\[ 1.6 = 1.56 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Ag^{+}]^{2}} \]
\[ 1.6 - 1.56 = -\frac{0.059}{2} \log \frac{1}{[Ag^{+}]^{2}} \]
\[ 0.04 = -\frac{0.059}{2} \times (-2 \log [Ag^{+}]) \]
\[ 0.04 = 0.059 \log [Ag^{+}] \]
\[ \log [Ag^{+}] = \frac{0.04}{0.059} = \frac{4}{5.9} \]
Step 4: Final Answer:
The value of \(\log_{10}[Ag^{+}]\) is \(\dfrac{4}{5.9}\).