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: Write the cell reaction and calculate standard emf.
The given half-cells are:
\[ \mathrm{Ag^+ + e^- \rightarrow Ag} \qquad E^\circ = 0.8 \, \mathrm{V} \] \[ \mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \qquad E^\circ = -0.76 \, \mathrm{V} \] Since silver has a higher reduction potential, it acts as the cathode and zinc acts as the anode. Therefore, the overall cell reaction is:
\[ \mathrm{Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)} \] Now, the standard emf of the cell is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.8 - (-0.76) = 1.56 \, \mathrm{V} \]
Step 2: Apply the Nernst equation.
For the reaction \[ \mathrm{Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)} \] the number of electrons transferred is \( n = 2 \).
The reaction quotient is: \[ Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Ag^+}]^2} \] Given that \( [\mathrm{Zn^{2+}}] = 1 \, \mathrm{M} \), we get: \[ Q = \frac{1}{[\mathrm{Ag^+}]^2} \] Now use the Nernst equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log Q \] \[ 1.6 = 1.56 - \frac{0.059}{2}\log \left( \frac{1}{[\mathrm{Ag^+}]^2} \right) \]
Step 3: Simplify the logarithmic term.
We know that: \[ \log \left( \frac{1}{[\mathrm{Ag^+}]^2} \right) = -2 \log [\mathrm{Ag^+}] \] Substituting this into the equation: \[ 1.6 = 1.56 - \frac{0.059}{2} \left( -2 \log [\mathrm{Ag^+}] \right) \] \[ 1.6 = 1.56 + 0.059 \log [\mathrm{Ag^+}] \] Now subtract \( 1.56 \) from both sides: \[ 1.6 - 1.56 = 0.059 \log [\mathrm{Ag^+}] \] \[ 0.04 = 0.059 \log [\mathrm{Ag^+}] \]
Step 4: Find the value of \( \log_{10}[\mathrm{Ag^+}] \).
\[ \log [\mathrm{Ag^+}] = \frac{0.04}{0.059} \] \[ \log [\mathrm{Ag^+}] = \frac{4}{5.9} \] Hence, \[ \log_{10}[\mathrm{Ag^+}] = \frac{4}{5.9} \] So, the correct option is \( (B) \).
Final Answer: \( \log_{10}[\mathrm{Ag^+}] = \dfrac{4}{5.9} \).